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Science:Math Exam Resources/Courses/MATH101/April 2014/Question 09 (b)/Solution 1

2015-04-09T19:25:51Z

ThomasWong:

Derive the formula in (a):

<math>\begin{align}
\sum_{n=0}^{\infty}n^2 x^{n-1}
&=\frac{d}{dx}\sum_{n=0}^{\infty} nx^n \\
&=\frac{d}{dx}\left(\frac{x}{(1-x)^2}\right) \\
&=\frac{1}{(1-x)^2}+\frac{2x}{(1-x)^3} \\
&=\frac{1+x}{(1-x)^3}.\end{align}</math>

Then, multiply <math>x</math>:

<math>\begin{align}
\sum_{n=0}^{\infty}n^2 x^{n}=\frac{x(1+x)}{(1-x)^3}.\end{align}</math>

Also, since

<math>\begin{align}
\lim_{n\to\infty}\left|\frac{(n+1)^2 x^n}{n^2 x^{n-1}}\right|
=\left(\lim_{n\to\infty}\left|\frac{(n+1)^2}{n^2}\right|\right)|x|=|x|,\end{align}</math>

by the ratio test, if <math>|x|<1</math>, the series converges and if <math>|x|>1</math>, the series diverges. Now check the boundaries. If <math>x=1</math>, the series diverges. If <math>x=-1</math>, the series become

<math>\begin{align}
\sum_{n=0}^{\infty}(-1)^{n}n^2
&=-1^2+2^2-3^2+4^2-5^2+6^2+\cdots
=(-1^2+2^2)+(-3^2+4^2)+(-5^2+6^2)+\cdots\\
&=(2-1)(1+2)+(3-2)(3+4)+(6-5)(5+6)+\cdots
=\sum_{n=0}^{\infty} n,\end{align}</math>

and hence diverges. Thus, for <math>x\in (-1,1)</math>, the series converges.

Science:Math Exam Resources/Courses/MATH110/April 2014/Question 01 (a)/Statement

2015-03-05T00:42:21Z

ThomasWong: Created page with "If <math>f</math> is defined on the closed interval <math>[-1,1]</math>, then <math>f</math> attains a global maximum on that interval."

If <math>f</math> is defined on the closed interval <math>[-1,1]</math>, then <math>f</math> attains a global maximum on that interval.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 05 (a)/Solution 1

2014-11-17T06:57:16Z

ThomasWong:

We evaluate these values by following three steps:

# Reduce the parameter to be in the interval <math>[0,2\pi)</math>.
# Apply the relevant special triangle.
# Check the sign of the result by remembering the <math>\cos(t)</math> gives the <math>x</math>-coordinate and <math>\sin(t)</math> gives the <math>y</math>-coordinate.

<ul>
<li><math>\sin\left(\frac{5\pi}{3}\right)</math>.
<ol>
<li>The angle <math>\frac{5\pi}{3}</math> is in the desired interval.</li>
<li>The angle <math>\frac{5\pi}{3}</math> is <math>\frac{\pi}{3}</math> below the <math>x</math>-axis. Applying the special triangle, we get <math>\sin\left(\frac{5\pi}{3}\right) =\pm \frac{\sqrt{3}}{2}</math>.</li>
<li>We are in the fourth quadrant so the <math>y</math>-coordinate is negative.</li></ol>

We have: <math>\sin\left(\frac{5\pi}{3}\right) =- \frac{\sqrt{3}}{2}</math>.</li>
<li><math>\cos\left(-\frac{\pi}{3}\right)</math>
<ol>
<li>The angle <math>-\frac{\pi}{3}</math> is not in the desired interval. So we apply periodicity:
<math>\begin{align}
\cos\left(-\frac{\pi}{3}\right) &= \cos\left(-\frac{\pi}{3} + 2\pi\right) \\
&= \cos\left(-\frac{\pi}{3}+\frac{6\pi}{3}\right)\\
&= \cos\left(\frac{5\pi}{3}\right)\end{align}</math></li>
<li>The angle <math>\frac{5\pi}{3}</math> is <math>\frac{\pi}{3}</math> below the <math>x</math>-axis. Applying the special triangle, we get <math>\cos\left(\frac{5\pi}{3}\right) =\pm \frac{1}{2}</math>.</li>
<li>We are in the fourth quadrant so the <math>x</math>-coordinate is positive.</li></ol>

We have: <math>\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}.</math></li>
<li><math>\tan\left(\frac{11\pi}{3}\right)</math>.
<ol>
<li>The angle <math>\frac{11\pi}{3}</math> is not in the desired interval. So we apply periodicity:
<math>\begin{align}
\tan\left(\frac{11\pi}{3}\right) &= \tan\left(\frac{11\pi}{3} - 2\pi\right) \\
&= \tan\left(\frac{11\pi}{3} - \frac{6\pi}{3}\right)\\
&= \tan\left(\frac{5\pi}{3}\right)\end{align}</math></li>
<li>The angle <math>\frac{5\pi}{3}</math> is <math>\frac{\pi}{3}</math> below the <math>x</math>-axis. Applying the special triangle, we get <math>\tan\left(\frac{5\pi}{3}\right) =\pm \sqrt{3}</math>.</li>
<li>We are in the fourth quadrant and since <math>\tan(t) = \frac{\sin(t)}{\cos(t)}</math>, we get that <math>\tan(t)</math> is negative.</li></ol>

Alternative, we can make use of the previous parts to get:

<math>\tan\left(\frac{11\pi}{3}\right) = \tan\left(\frac{5\pi}{3}\right) = \frac{\sin\left(\frac{11\pi}{3}\right)}{\cos\left(\frac{11\pi}{3}\right)} = \frac{\frac{-\sqrt{3}}{2}}{\frac{1}{2}}</math>

Either way, we have: <math>\tan\left(\frac{11\pi}{3}\right) = -\sqrt{3}</math></li></ul>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 05 (a)/Solution 1

2014-11-17T06:56:09Z

ThomasWong:

We evaluate these values by following three steps:

# Reduce the parameter to be in the interval <math>[0,2\pi)</math>.
# Apply the relevant special triangle.
# Check the sign of the result by remembering the <math>\cos(t)</math> gives the <math>x</math>-coordinate and <math>\sin(t)</math> gives the <math>y</math>-coordinate.

<ul>
<li><math>\sin\left(\frac{5\pi}{3}\right)</math>.
<ol>
<li>The angle <math>\frac{5\pi}{3}</math> is in the desired interval.</li>
<li>The angle <math>\frac{5\pi}{3}</math> is <math>\frac{\pi}{3}</math> below the <math>x</math>-axis. Applying the special triangle, we get <math>\sin\left(\frac{5\pi}{3}\right) =\pm \frac{\sqrt{3}}{2}</math>.</li>
<li>We are in the fourth quadrant so the <math>y</math>-coordinate is negative.</li></ol>

We have: <math>\sin\left(\frac{5\pi}{3}\right) =- \frac{\sqrt{3}}{2}</math>.</li>
<li><math>\cos\left(-\frac{\pi}{3}\right)</math>
<ol>
<li>The angle <math>-\frac{\pi}{3}</math> is not in the desired interval. So we apply periodicity:
<math>\begin{align}
\cos\left(-\frac{\pi}{3}\right) &= \cos\left(-\frac{\pi}{3} + 2\pi\right) \\
&= \cos\left(-\frac{\pi}{3}+\frac{6\pi}{3}\right)\\
&= \cos\left(\frac{5\pi}{3}\right)\end{align}</math></li>
<li>The angle <math>\frac{5\pi}{3}</math> is <math>\frac{\pi}{3}</math> below the <math>x</math>-axis. Applying the special triangle, we get <math>\cos\left(\frac{5\pi}{3}\right) =\pm \frac{1}{2}</math>.</li>
<li>We are in the fourth quadrant so the <math>x</math>-coordinate is positive.</li></ol>

We have: <math>\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}.</math></li>
<li><math>\tan\left(\frac{11\pi}{3}\right)</math>.
<ol>
<li>The angle <math>\frac{11\pi}{3}</math> is not in the desired interval. So we apply periodicity:
<math>\begin{align}
\tan\left(\frac{11\pi}{3}\right) &= \tan\left(\frac{11\pi}{3} - 2\pi\right) \\
&= \tan\left(\frac{11\pi}{3} - \frac{6\pi}{3}\right)\\
&= \tan\left(\frac{5\pi}{3}\right)\end{align}</math></li>
<li>The angle <math>\frac{5\pi}{3}</math> is <math>\frac{\pi}{3}</math> below the <math>x</math>-axis. Applying the special triangle, we get <math>\tan\left(\frac{5\pi}{3}\right) =\pm \sqrt{3}</math>.</li>
<li>We are in the fourth quadrant and since <math>\tan(t) = \frac{\sin(t)}{\cos(t)}</math>, we get that <math>\tan(t)</math> is negative.</li></ol>

Alternative, we can make use of the previous parts to get: <math>\tan\left(\frac{11\pi}{3}\right) = \tan\left(\frac{5\pi}{3}\right) = \frac{\sin\left(\frac{11\pi}{3}\right)}{\cos\left(\frac{11\pi}{3}\right)} = \frac{\frac{-\sqrt{3}}{2}}{\frac{1}{2}}</math> Either way, we have: <math>\tan\left(\frac{11\pi}{3}\right) = -\sqrt{3}</math></li></ul>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 05 (a)/Solution 1

2014-11-17T06:55:13Z

ThomasWong:

We evaluate these values by following three steps:

# Reduce the parameter to be in the interval <math>[0,2\pi)</math>.
# Apply the relevant special triangle.
# Check the sign of the result by remembering the <math>\cos(t)</math> gives the <math>x</math>-coordinate and <math>\sin(t)</math> gives the <math>y</math>-coordinate.

<ul>
<li><math>\sin\left(\frac{5\pi}{3}\right)</math>.
<ol>
<li>The angle <math>\frac{5\pi}{3}</math> is in the desired interval.</li>
<li>The angle <math>\frac{5\pi}{3}</math> is <math>\frac{\pi}{3}</math> below the <math>x</math>-axis. Applying the special triangle, we get <math>\sin\left(\frac{5\pi}{3}\right) =\pm \frac{\sqrt{3}}{2}</math>.</li>
<li>We are in the fourth quadrant so the <math>y</math>-coordinate is negative.</li></ol>

We have: <math>\sin\left(\frac{5\pi}{3}\right) =- \frac{\sqrt{3}}{2}</math>.</li>
<li><math>\cos\left(-\frac{\pi}{3}\right)</math>
<ol>
<li>The angle <math>-\frac{\pi}{3}</math> is not in the desired interval. So we apply periodicity:
<math>\begin{align}
\cos\left(-\frac{\pi}{3}\right) &= \cos\left(-\frac{\pi}{3} + 2\pi\right) \\
= \cos\left(-\frac{\pi}{3}+\frac{6\pi}{3}\right)\\
= \cos\left(\frac{5\pi}{3}\right)\end{align}</math></li>
<li>The angle <math>\frac{5\pi}{3}</math> is <math>\frac{\pi}{3}</math> below the <math>x</math>-axis. Applying the special triangle, we get <math>\cos\left(\frac{5\pi}{3}\right) =\pm \frac{1}{2}</math>.</li>
<li>We are in the fourth quadrant so the <math>x</math>-coordinate is positive.</li></ol>

We have: <math>\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}.</math></li>
<li><math>\tan\left(\frac{11\pi}{3}\right)</math>.
<ol>
<li>The angle <math>\frac{11\pi}{3}</math> is not in the desired interval. So we apply periodicity:
<math>\begin{align}
\tan\left(\frac{11\pi}{3}\right) &= \tan\left(\frac{11\pi}{3} - 2\pi\right) \\
= \tan\left(\frac{11\pi}{3} - \frac{6\pi}{3}\right)\\
= \tan\left(\frac{5\pi}{3}\right)\end{align}</math></li>
<li>The angle <math>\frac{5\pi}{3}</math> is <math>\frac{\pi}{3}</math> below the <math>x</math>-axis. Applying the special triangle, we get <math>\tan\left(\frac{5\pi}{3}\right) =\pm \sqrt{3}</math>.</li>
<li>We are in the fourth quadrant and since <math>\tan(t) = \frac{\sin(t)}{\cos(t)}</math>, we get that <math>\tan(t)</math> is negative.</li></ol>

Alternative, we can make use of the previous parts to get: <math>\tan\left(\frac{11\pi}{3}\right) = \tan\left(\frac{5\pi}{3}\right) = \frac{\sin\left(\frac{11\pi}{3}\right)}{\cos\left(\frac{11\pi}{3}\right)} = \frac{\frac{-\sqrt{3}}{2}}{\frac{1}{2}}</math> Either way, we have: <math>\tan\left(\frac{11\pi}{3}\right) = -\sqrt{3}</math></li></ul>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 05 (a)/Solution 1

2014-11-17T06:54:25Z

ThomasWong:

We evaluate these values by following three steps:

# Reduce the parameter to be in the interval <math>[0,2\pi)</math>.
# Apply the relevant special triangle.
# Check the sign of the result by remembering the <math>\cos(t)</math> gives the <math>x</math>-coordinate and <math>\sin(t)</math> gives the <math>y</math>-coordinate.

<ul>
<li><math>\sin\left(\frac{5\pi}{3}\right)</math>.
<ol>
<li>The angle <math>\frac{5\pi}{3}</math> is in the desired interval.</li>
<li>The angle <math>\frac{5\pi}{3}</math> is <math>\frac{\pi}{3}</math> below the <math>x</math>-axis. Applying the special triangle, we get <math>\sin\left(\frac{5\pi}{3}\right) =\pm \frac{\sqrt{3}}{2}</math>.</li>
<li>We are in the fourth quadrant so the <math>y</math>-coordinate is negative.</li></ol>

We have: <math>\sin\left(\frac{5\pi}{3}\right) =- \frac{\sqrt{3}}{2}</math>.</li>
<li><math>\cos\left(-\frac{\pi}{3}\right)</math>
<ol>
<li>The angle <math>-\frac{\pi}{3}</math> is not in the desired interval. So we apply periodicity:
<math>\begin{aligned}
\cos\left(-\frac{\pi}{3}\right) &= \cos\left(-\frac{\pi}{3} + 2\pi\right) \\
= \cos\left(-\frac{\pi}{3}+\frac{6\pi}{3}\right)\\
= \cos\left(\frac{5\pi}{3}\right)\end{aligned}</math></li>
<li>The angle <math>\frac{5\pi}{3}</math> is <math>\frac{\pi}{3}</math> below the <math>x</math>-axis. Applying the special triangle, we get <math>\cos\left(\frac{5\pi}{3}\right) =\pm \frac{1}{2}</math>.</li>
<li>We are in the fourth quadrant so the <math>x</math>-coordinate is positive.</li></ol>

We have: <math>\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}.</math></li>
<li><math>\tan\left(\frac{11\pi}{3}\right)</math>.
<ol>
<li>The angle <math>\frac{11\pi}{3}</math> is not in the desired interval. So we apply periodicity:
<math>\begin{aligned}
\tan\left(\frac{11\pi}{3}\right) &= \tan\left(\frac{11\pi}{3} - 2\pi\right) \\
= \tan\left(\frac{11\pi}{3} - \frac{6\pi}{3}\right)\\
= \tan\left(\frac{5\pi}{3}\right)\end{aligned}</math></li>
<li>The angle <math>\frac{5\pi}{3}</math> is <math>\frac{\pi}{3}</math> below the <math>x</math>-axis. Applying the special triangle, we get <math>\tan\left(\frac{5\pi}{3}\right) =\pm \sqrt{3}</math>.</li>
<li>We are in the fourth quadrant and since <math>\tan(t) = \frac{\sin(t)}{\cos(t)}</math>, we get that <math>\tan(t)</math> is negative.</li></ol>

Alternative, we can make use of the previous parts to get: <math>\tan\left(\frac{11\pi}{3}\right) = \tan\left(\frac{5\pi}{3}\right) = \frac{\sin\left(\frac{11\pi}{3}\right)}{\cos\left(\frac{11\pi}{3}\right)} = \frac{\frac{-\sqrt{3}}{2}}{\frac{1}{2}}</math> Either way, we have: <math>\tan\left(\frac{11\pi}{3}\right) = -\sqrt{3}</math></li></ul>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 01 (c)/Solution 1

2014-11-17T00:13:21Z

ThomasWong:

'''False:''' To hunt for discontinuities, we have to check inside each piece and also at the boundary.

<ul>
<li>For the boundary to be continuous, we must have:
<math>\begin{align}
\lim_{x\rightarrow 1^-} f(x) &=\lim_{x\rightarrow 1^+} f(x)\\
\lim_{x\rightarrow 1^-} \frac{3}{x+2} &=\lim_{x\rightarrow 1^+} \sqrt{x}\\
\frac{3}{1+2} &=\sqrt{1} &{\rm Direct\; Substitution.}\\
1&=1.\end{align}</math>
So the limits exists and is equal to <math>1</math> which happens to be <math>f(1) =1</math>. So the function in continuous at the boundary.</li>
<li>On the right of <math>x=1</math>, the function is continuous since <math>\sqrt{x}</math> is only undefined for <math>x<0</math> which is not covered by this case.</li>
<li>On the left of <math>x=1</math>, the function is discontinuous at <math>x=-2</math> since the denominator is <math>0</math>. This fall inside the region considered. So <math>f(x)</math> is discontinuous at <math>x=-2</math>.</li></ul>

That means <math>f(x)</math> is NOT continuous over all real numbers.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 10 (a)/Solution 1

2014-11-16T06:37:46Z

ThomasWong: Created page with "We have <math>4</math> slots to fill and <math>7</math> people to fill them. So that means, we have <math>7</math> choices for the first slot, <math>6</math> for the second, <..."

We have <math>4</math> slots to fill and <math>7</math> people to fill them. So that means, we have <math>7</math> choices for the first slot, <math>6</math> for the second, <math>5</math> for the third and <math>4</math> for the fourth. Since the number of choices I have for each slot is not dependent on my choice of the previous spots. I will be multiplying these numbers. So the total number of ways to fill these slots is <math>7\cdot6\cdot5\cdot4 = \frac{7!}{3!} = 840</math>. There are <math>840</math> ways to fill the slots.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 10 (b)/Solution 1

2014-11-16T06:37:12Z

ThomasWong: Created page with "When selecting courses, the order does not matter. So I am really interested in a combination here. I have want to select <math>5</math> of the <math>9</math> available course..."

When selecting courses, the order does not matter. So I am really interested in a combination here. I have want to select <math>5</math> of the <math>9</math> available courses, so I am looking at <math>\binom{9}{5}</math>.

<math>\begin{align}
\binom{9}{5} &= \frac{9!}{5! (9-5)!}\\
&= \frac{9!}{5!}\cdot \frac{1}{4!}\\
&= \frac{9\cdot8\cdot7\cdot 6}{4\cdot3\cdot2\cdot1}\\
&= 9\cdot2\cdot7 &{\rm Simplify}\\
&=126\end{align}</math>

There are <math>126</math> ways to fill the 5 timeslots.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 10 (b)/Statement

2014-11-16T06:36:33Z

ThomasWong:

There are <math>9</math> courses I want to take next term. All <math>9</math> courses are offered during the same <math>5</math> time slots. How many ways do I have to fill those <math>5</math> time slots?

''For this question, a numerical answer is required as a final answer.''

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 10 (b)/Statement

2014-11-16T06:36:11Z

ThomasWong: Created page with "There are $9$ courses I want to take next term. All $9$ courses are offered during the same $5$ time slots. How many ways do I have to fill those $5$ time slots? \emph{For th..."

There are $9$ courses I want to take next term. All $9$ courses are offered during the same $5$ time slots. How many ways do I have to fill those $5$ time slots?

\emph{For this question, a numerical answer is required as a final answer.}

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 10 (a)/Statement

2014-11-16T06:35:41Z

ThomasWong: Created page with "How many ways can <math>7</math> graduate students be assigned to teach the <math>4</math> sections of MATH110 next year if each graduate student can be assigned to at most on..."

How many ways can <math>7</math> graduate students be assigned to teach the <math>4</math> sections of MATH110 next year if each graduate student can be assigned to at most one section?

''For this question, a numerical answer is required as a final answer.''

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 09/Solution 1

2014-11-16T06:34:30Z

ThomasWong:

We first check that the point <math>(0,\pi)</math> is on the line:

<math>\begin{align}
\sin(0+\pi) &= 0 \cdot \pi\\
\sin(\pi) &=0\\
0&=0\end{align}</math>

Check. The point is on the line. Next we want to find the derivative of the function. We apply implicit differentiation and differentiate both sides:

<math>\begin{align}
{\frac{{\rm d}}{{\rm d}x}} \sin(x+y) &= {\frac{{\rm d}}{{\rm d}x}} xy\\
\left(1+{\frac{{\rm d} {y}}{{\rm d}x}}\right)\cos(x+y) &= x{\frac{{\rm d} {y}}{{\rm d}x}} + y\end{align}</math>

The right side was obtained via the chain rule and the left hand side was obtained via a product rule.

<math>\begin{align}
\cos(x+y) + {\frac{{\rm d} {y}}{{\rm d}x}}\cos(x+y) &= x{\frac{{\rm d} {y}}{{\rm d}x}} + y\\
{\frac{{\rm d} {y}}{{\rm d}x}}\cos(x+y) - x{\frac{{\rm d} {y}}{{\rm d}x}} &= y - \cos(x+y) \\
{\frac{{\rm d} {y}}{{\rm d}x}}\left( \cos(x+y) - x\right)&= y - \cos(x+y) \\
{\frac{{\rm d} {y}}{{\rm d}x}}&= \frac{y - \cos(x+y)}{\cos(x+y) - x} \\\end{align}</math>

substitute

<math>\begin{align}
{\frac{{\rm d} {y}}{{\rm d}x}}&= \frac{\pi - \cos(0+\pi)}{\cos(0+\pi) - 0} \\
{\frac{{\rm d} {y}}{{\rm d}x}}&= \frac{\pi - (-1) }{(-1) - 0} \\
{\frac{{\rm d} {y}}{{\rm d}x}}&= \frac{\pi + 1}{-1} = -(\pi+1)\\\end{align}</math>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 09/Solution 1

2014-11-16T06:33:53Z

ThomasWong: Created page with "We first check that the point <math>(0,\pi)</math> is on the line: <math>\begin{aligned} \sin(0+\pi) &= 0 \cdot \pi\\ \sin(\pi) &=0\\ 0&=0\end{aligned}</math> Check. The poi..."

We first check that the point <math>(0,\pi)</math> is on the line:

<math>\begin{aligned}
\sin(0+\pi) &= 0 \cdot \pi\\
\sin(\pi) &=0\\
0&=0\end{aligned}</math>

Check. The point is on the line. Next we want to find the derivative of the function. We apply implicit differentiation and differentiate both sides:

<math>\begin{align}
{\frac{{\rm d}}{{\rm d}x}} \sin(x+y) &= {\frac{{\rm d}}{{\rm d}x}} xy\\
\left(1+{\frac{{\rm d} {y}}{{\rm d}x}}\right)\cos(x+y) &= x{\frac{{\rm d} {y}}{{\rm d}x}} + y\end{align}</math>

The right side was obtained via the chain rule and the left hand side was obtained via a product rule.

<math>\begin{align}
\cos(x+y) + {\frac{{\rm d} {y}}{{\rm d}x}}\cos(x+y) &= x{\frac{{\rm d} {y}}{{\rm d}x}} + y\\
{\frac{{\rm d} {y}}{{\rm d}x}}\cos(x+y) - x{\frac{{\rm d} {y}}{{\rm d}x}} &= y - \cos(x+y) \\
{\frac{{\rm d} {y}}{{\rm d}x}}\left( \cos(x+y) - x\right)&= y - \cos(x+y) \\
{\frac{{\rm d} {y}}{{\rm d}x}}&= \frac{y - \cos(x+y)}{\cos(x+y) - x} \\\end{align}</math>

substitute

<math>\begin{align}
{\frac{{\rm d} {y}}{{\rm d}x}}&= \frac{\pi - \cos(0+\pi)}{\cos(0+\pi) - 0} \\
{\frac{{\rm d} {y}}{{\rm d}x}}&= \frac{\pi - (-1) }{(-1) - 0} \\
{\frac{{\rm d} {y}}{{\rm d}x}}&= \frac{\pi + 1}{-1} = -(\pi+1)\\\end{align}</math>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 09/Statement

2014-11-16T06:30:28Z

ThomasWong: Created page with "Find the derivative of <math>y</math> with respect to <math>x</math> at the point <math>P = (0,\pi)</math> along the curve: <math>\sin(x+y) = xy</math>"

Find the derivative of <math>y</math> with respect to <math>x</math> at the point <math>P = (0,\pi)</math> along the curve:

<math>\sin(x+y) = xy</math>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 08/Solution 1

2014-11-16T06:29:11Z

ThomasWong:

The condition for <math>f</math> to be continuous is

<math>\begin{align}
\lim_{x\rightarrow t^-} f(x) &= \lim_{x\rightarrow t^+} f(x)\\
e^{-t}&=2(t+1)\end{align}</math>

Since both piece of the function are continuous, we can use direct substitution and find <math>t</math>. We must show there exists a solution to the above equation. Equivalently we want to show that the following function <math>g(t)</math> has a zero

<math>\begin{align}
g(t)=e^{-t}-2(t+1)\end{align}</math>

Note that <math>g</math> is continuous since it is constructed from continuous functions. We aim to apply IVT. Observe

<math>\begin{align}
g(0)&=e^0-2<0\\
g(-1)&=e^1-2(-1+1)=e>0\end{align}</math>

Hence by the IVT there exists <math>c \in (-1,0)</math> such that <math>g(c)=0</math>. At <math>t=c</math>, we have <math>e^{-c} = 2(c+1)</math>. In turn, this means we have the two one sided limits equalling each other and hence <math>f(x)</math> will be continuous by choosing <math>x=c</math>.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 08/Solution 1

2014-11-16T06:28:34Z

ThomasWong: Created page with "The condition for <math>f</math> to be continuous is <math>\begin{aligned} \lim_{x\rightarrow t^-} f(x) &= \lim_{x\rightarrow t^+} f(x)\\ e^{-t}&=2(t+1)\end{aligned}</math> ..."

The condition for <math>f</math> to be continuous is

<math>\begin{aligned}
\lim_{x\rightarrow t^-} f(x) &= \lim_{x\rightarrow t^+} f(x)\\
e^{-t}&=2(t+1)\end{aligned}</math>

Since both piece of the function are continuous, we can use direct substitution and find <math>t</math>. We must show there exists a solution to the above equation. Equivalently we want to show that the following function <math>g(t)</math> has a zero

<math>\begin{aligned}
g(t)=e^{-t}-2(t+1)\end{aligned}</math>

Note that <math>g</math> is continuous since it is constructed from continuous functions. We aim to apply IVT. Observe

<math>\begin{aligned}
g(0)&=e^0-2<0\\
g(-1)&=e^1-2(-1+1)=e>0\end{aligned}</math>

Hence by the IVT there exists <math>c \in (-1,0)</math> such that <math>g(c)=0</math>. At <math>t=c</math>, we have <math>e^{-c} = 2(c+1)</math>. In turn, this means we have the two one sided limits equalling each other and hence <math>f(x)</math> will be continuous by choosing <math>x=c</math>.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 08/Statement

2014-11-16T06:28:07Z

ThomasWong: Created page with "Let <math>f(x) = \left\{ \begin{array}{lr} e^{-x} & \text{if}\quad x < t\\ 2(x+1) & \text{if} \quad x \geq t \end{array} \right.</math> Explain why ..."

Let <math>f(x) = \left\{
\begin{array}{lr}
e^{-x} & \text{if}\quad x < t\\
2(x+1) & \text{if} \quad x \geq t
\end{array}
\right.</math> Explain why there exists a number <math>t</math> such that <math>f</math> is continuous.

(Hint: apply the Intermediate Value Theorem to the function <math>e^{-t}-2(t+1)</math>.)

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 07/Solution 1

2014-11-16T06:26:44Z

ThomasWong: Created page with "<ul> <li>We can solve this problem ‘normally’ by using product and chain rule or by using logarithmic differentiation. <math>\begin{align} f'(x)&=7(x^2+1)^6(x^4+..."

<ul>
<li>We can solve this problem ‘normally’ by using product and chain rule or by using logarithmic differentiation.
<math>\begin{align}
f'(x)&=7(x^2+1)^6(x^4+2)^5(x^6+3)^3(x^8+4)\cdot2x + 5(x^2+1)^7(x^4+2)^4(x^6+3)^3(x^8+4)\cdot4x^3\\
&\quad + 3(x^2+1)^7(x^4+2)^5(x^6+3)^2(x^8+4)\cdot 6x^5 + (x^2+1)^7(x^4+2)^5(x^6+3)^3\cdot 8x^7\end{align}</math>
Note that we made use of the general product rule:
<math>\begin{align}
\frac{d}{dx}(f_1 f_2 f_3 f_4)&= \frac{df_1}{dx} f_2 f_3 f_4 + f_1 \frac{df_2}{dx} f_3 f_4 + f_1 f_2 \frac{df_3}{dx} f_4 + f_1 f_2 f_3 \frac{df_4}{dx}. \end{align}</math></li>
<li>Alternatively, let <math>y=f(x)</math> and
<math>\begin{align}
\ln y &= \ln \left((x^2+1)^7(x^4+2)^5(x^6+3)^3(x^8+4)\right)\\
&= 7 \ln (x^2+1) + 5 \ln (x^4+2) + 3 \ln (x^6+3) + \ln (x^8+4)\end{align}</math>
Differentiating implicitly yields
<math>\begin{align}
\frac{1}{y} \frac{dy}{dx}&= \frac{7} { (x^2+1)}\cdot 2x + \frac{5}{(x^4+2)}\cdot 4x^3 + \frac{3} {(x^6+3)}\cdot 6x^5 + \frac{1} {(x^8+4)} \cdot 8x^7\end{align}</math>
So
<math>\begin{align}
f'(x)&= (x^2+1)^7(x^4+2)^5(x^6+3)^3(x^8+4) \left( \frac{14x} { (x^2+1)} + \frac{20x^3}{(x^4+2)} + \frac{18x^5} {(x^6+3)} + \frac{8x^7} {(x^8+4)} \right)\end{align}</math>
which gives the same result.</li></ul>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 07/Statement

2014-11-16T06:25:20Z

ThomasWong: Created page with "Find the derivative of the function <math>f(x)=(x^2+1)^7(x^4+2)^5(x^6+3)^3(x^8+4).</math> (There is no need to simplify your answer.)"

Find the derivative of the function <math>f(x)=(x^2+1)^7(x^4+2)^5(x^6+3)^3(x^8+4).</math> (There is no need to simplify your answer.)

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 06 (c)/Solution 1

2014-11-16T06:24:02Z

ThomasWong: Created page with "Recall that <math>f'(x) = \left(3x^2 + 4x + 1\right)\cdot e^{x^3 + 2x^2 + x - 3}.</math> From part a), we know that <math>e^{x^3 + 2x^2 + x - 3}\neq 0</math>. So that means ..."

Recall that

<math>f'(x) = \left(3x^2 + 4x + 1\right)\cdot e^{x^3 + 2x^2 + x - 3}.</math>

From part a), we know that <math>e^{x^3 + 2x^2 + x - 3}\neq 0</math>. So that means <math>f'(x) = 0</math> when <math>3x^2 + 4x + 1=0</math>. Factoring, we get <math>(3x+1)(x+1)=0</math>, which means <math>x=-\frac{1}{3},-1</math>. That means there are two solutions for when <math>f'(x)=0</math>.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 06 (c)/Statement

2014-11-16T06:23:34Z

ThomasWong: Created page with "Consider the function: <math>f(x) = \dfrac{e^{\left(x^3\right)}e^{\left(2x^2\right)}e^{x}}{e^3}.</math> How many solutions are there to the equation <math>f'(x)=0</math>? Just..."

Consider the function: <math>f(x) = \dfrac{e^{\left(x^3\right)}e^{\left(2x^2\right)}e^{x}}{e^3}.</math> How many solutions are there to the equation <math>f'(x)=0</math>? Justify your answer.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 06 (b)/Solution 1

2014-11-16T06:22:31Z

ThomasWong:

<ul>
<li>Since we have simplified, we can simply apply the chain rule to it using <math>u = x^3 + 2x^2 + x - 3</math>.
<math>\begin{align}
f(u) &= e^u & f'(u) &= e^u\\
u(x) & x^3 + 2x^2 + x - 3 & u'(x) &= 3x^2 + 4x + 1.\end{align}</math>
Combining, we get:
<math>f'(x) = \left(3x^2 + 4x + 1\right)\cdot e^{x^3 + 2x^2 + x - 3}</math></li>
<li>Alternatively, we can start with the initial function and use log-diff:
<math>\begin{align}
f(x) &= \frac{e^{\left(x^3\right)}e^{\left(2x^2\right)}e^{x}}{e^3}\\
\ln(f(x)) &= \ln\left(\frac{e^{\left(x^3\right)}e^{\left(2x^2\right)}e^{x}}{e^3}\right)\\
&= \ln\left(e^{\left(x^3\right)}\right) + \ln\left(e^{\left(2x^2\right)}\right) + \ln\left(e^{x}\right) - \ln\left({e^3}\right)\\
&= x^3 \ln\left(e\right) + 2x^2 \ln\left(e\right) +x \ln\left(e\right) - 3\ln\left(e\right)\\
\ln(f(x))&= x^3 + 2x^2 +x - 3\\
{\frac{{\rm d}}{{\rm d}x}} \ln(f(x)) &= {\frac{{\rm d}}{{\rm d}x}} \left(x^3 + 2x^2 +x - 3\right)\\
\frac{f'(x)}{f(x)} &= 3x^2 + 4x +1\\
f'(x) &= \left(3x^2 + 4x +1\right) f(x)\\
f'(x) &= \left(3x^2 + 4x +1\right)]\cdot \frac{e^{\left(x^3\right)}e^{\left(2x^2\right)}e^{x}}{e^3}\\\end{align}</math></li></ul>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 06 (b)/Solution 1

2014-11-16T06:21:52Z

ThomasWong:

<ul>
<li>Since we have simplified, we can simply apply the chain rule to it using <math>u = x^3 + 2x^2 + x - 3</math>.
<math>\begin{aligned}
f(u) &= e^u & f'(u) &= e^u\\
u(x) & x^3 + 2x^2 + x - 3 & u'(x) &= 3x^2 + 4x + 1.\end{aligned}</math>
Combining, we get:
<math>f'(x) = \left(3x^2 + 4x + 1\right)\cdot e^{x^3 + 2x^2 + x - 3}</math></li>
<li>Alternatively, we can start with the initial function and use log-diff:
<math>\begin{aligned}
f(x) &= \frac{e^{\left(x^3\right)}e^{\left(2x^2\right)}e^{x}}{e^3}\\
\ln(f(x)) &= \ln\left(\frac{e^{\left(x^3\right)}e^{\left(2x^2\right)}e^{x}}{e^3}\right)\\
&= \ln\left(e^{\left(x^3\right)}\right) + \ln\left(e^{\left(2x^2\right)}\right) + \ln\left(e^{x}\right) - \ln\left({e^3}\right)\\
&= x^3 \ln\left(e\right) + 2x^2 \ln\left(e\right) +x \ln\left(e\right) - 3\ln\left(e\right)\\
\ln(f(x))&= x^3 + 2x^2 +x - 3\\
{\frac{{\rm d}}{{\rm d}x}} \ln(f(x)) &= {\frac{{\rm d}}{{\rm d}x}} \left(x^3 + 2x^2 +x - 3\right)\\
\frac{f'(x)}{f(x)} &= 3x^2 + 4x +1\\
f'(x) &= \left(3x^2 + 4x +1\right) f(x)\\
f'(x) &= \left(3x^2 + 4x +1\right)]\cdot \frac{e^{\left(x^3\right)}e^{\left(2x^2\right)}e^{x}}{e^3}\\\end{aligned}</math></li></ul>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 06 (b)/Solution 1

2014-11-16T06:20:16Z

ThomasWong: Blanked the page

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 06 (b)/Statement

2014-11-16T06:17:42Z

ThomasWong:

Consider the function: <math>f(x) = \dfrac{e^{\left(x^3\right)}e^{\left(2x^2\right)}e^{x}}{e^3}.</math> Find the derivative of <math>f(x)</math>.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 06 (b)/Solution 1

2014-11-16T06:16:04Z

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 06 (b)/Statement

2014-11-16T06:15:36Z

Consider the function: <math>f(x) = \dfrac{e^{\left(x^3\right)}e^{\left(2x^2\right)}e^{x}}{e^3}.</math> How many solutions are there to the equation <math>f'(x)=0</math>? Justify your answer.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 06 (a)/Solution 1

2014-11-16T06:14:38Z

ThomasWong: Created page with "We first simplify using the exponential rules to get <math>f(x) = e^{x^3 + 2x^2 + x - 3}.</math> Then we can solve for <math>f(x) = 0</math>. <math>\begin{align} e^{x^3 + 2x^..."

We first simplify using the exponential rules to get <math>f(x) = e^{x^3 + 2x^2 + x - 3}.</math> Then we can solve for <math>f(x) = 0</math>.

<math>\begin{align}
e^{x^3 + 2x^2 + x - 3} &= 0\\
x^3 + 2x^2 + x - 3 &= \ln(0).\end{align}</math>

By taking the natural logarithm of both sides, we get a <math>\ln(0)</math> on the right hand side which is bad since <math>0</math> is not in the domain of the natural logarithm. So we conclude that there cannot be a solution to the equation <math>f(x)=0</math>. Hence, we have <math>0</math> solutions.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 06 (a)/Statement

2014-11-16T06:13:22Z

Consider the function: <math>f(x) = \dfrac{e^{\left(x^3\right)}e^{\left(2x^2\right)}e^{x}}{e^3}.</math> How many solutions are there to the equation <math>f(x)=0</math>? Justify your answer.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 05 (b)/Solution 1

2014-11-15T01:27:51Z

ThomasWong: Created page with "To find the equation of the tangent line, we need to find the slope via the derivative. We can either apply the quotient rule or the power rule (by recognising <math>y = \tan(..."

To find the equation of the tangent line, we need to find the slope via the derivative. We can either apply the quotient rule or the power rule (by recognising <math>y = \tan(x)^{-1}</math>. Using the quotient rule:

<math>\begin{align}
u(x) &= \cos(x) & u'(x) = -\sin(x)\\
v(x) &= \sin(x) & v'(x) = \cos(x)\end{align}</math>

This gives:

<math>\begin{align}
{\frac{{\rm d} {y}}{{\rm d}x}} &= \frac{v u' - u v'}{v^2}\\
&= \frac{\sin(x) \cdot (-\sin(x) - \cos(x) \cdot \cos(x)}{[\sin(x)]^2}\\
&= \frac{- \sin^2(x) - \cos^2(x) }{\sin^2(x)}\\
&= \frac{- \left( \sin^2(x) + \cos^2(x)\right) }{\sin^2(x)}\\
&= \frac{-1}{\sin^2(x)}\\\end{align}</math>

At <math>x =\frac{5\pi}{3}</math>, we have <math>\sin\left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2}</math>, So that means the slope of the curve at <math>x =\frac{5\pi}{3}</math> is

<math>\begin{align}
y' &= \frac{-1}{\sin^2\left(x\right)}\\
&= \frac{-1}{\sin^2\left(\frac{5\pi}{3}\right)}\\
&= \frac{-1}{\left[-\frac{\sqrt{3}}{2}\right]^2}\\
&= \frac{-1}{\frac{3}{4}}\\
&= \frac{-4}{3}\\\end{align}</math>

We will also need the <math>y</math>-value at the point

<math>\begin{align}
y &= \frac{\cos(x)}{\sin(x)}\\
&= \frac{1}{\tan\left(\frac{5\pi}{3}\right)}\\
&= -\frac{1}{\tan\left(\frac{\pi}{3}\right)}\\
&= -\frac{1}{\sqrt{3}}\\
&= -\frac{\sqrt{3}}{3}\end{align}</math>

So that means our tangent line is:

<math>y = -\frac{\sqrt{3}}{3} - \frac{4}{3}\left(x - \frac{5\pi}{3}\right)</math>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 05 (b)/Statement

2014-11-15T01:26:00Z

ThomasWong: Created page with "Find the equation of the line tangent to the curve <math>y=\dfrac{\cos x}{\sin x}</math> at <math>x=\dfrac{5\pi}{3}</math>."

Find the equation of the line tangent to the curve <math>y=\dfrac{\cos x}{\sin x}</math> at <math>x=\dfrac{5\pi}{3}</math>.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 05 (a)/Solution 1

2014-11-15T01:23:13Z

ThomasWong: Created page with "* <math>\sin\left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2}.</math> * <math>\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}.</math> * <math>\tan\left(\frac{11\pi}{3}\right) = ..."

* <math>\sin\left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2}.</math>
* <math>\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}.</math>
* <math>\tan\left(\frac{11\pi}{3}\right) = -\sqrt{3}.</math>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 05 (a)/Statement

2014-11-15T01:22:35Z

ThomasWong: Created page with "Evaluate the following three quantities: <math>\sin\left(\frac{5\pi}{3}\right)</math>, <math>\cos\left(-\frac{\pi}{3}\right)</math>, <math>\tan\left(\frac{11\pi}{3}\right)</ma..."

Evaluate the following three quantities: <math>\sin\left(\frac{5\pi}{3}\right)</math>, <math>\cos\left(-\frac{\pi}{3}\right)</math>, <math>\tan\left(\frac{11\pi}{3}\right)</math>.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 04 (c)/Solution 1

2014-11-15T01:20:01Z

ThomasWong: Created page with "We want to apply the chain rule. Set <math>u = \sqrt{x}-1</math>. <math>\begin{align} g(u) &= \sqrt{u}-1 & g'(u) &= \frac{1}{2\sqrt{u}}\\ u(x) &= \sqrt{x}-1 & u'(x) &= \frac{..."

We want to apply the chain rule. Set <math>u = \sqrt{x}-1</math>.

<math>\begin{align}
g(u) &= \sqrt{u}-1 & g'(u) &= \frac{1}{2\sqrt{u}}\\
u(x) &= \sqrt{x}-1 & u'(x) &= \frac{1}{2\sqrt{x}}\end{align}</math>

Combining, we get:

<math>\begin{align}
g'(x) &= \frac{1}{2\sqrt{x}} \cdot \frac{1}{2\sqrt{\sqrt{x}-1}}\\
&= \frac{1}{4\sqrt{x}\cdot\sqrt{\sqrt{x}-1}}\\
&= \frac{1}{4\sqrt{x \left(\sqrt{x}-1\right)}}\end{align}</math>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 04 (c)/Statement

2014-11-15T01:19:26Z

ThomasWong: Created page with "Let <math>f(x)=\sqrt{x}-1</math> and <math>g(x)=(f\circ f)(x)</math>. Find the derivative of <math>g(x)</math>."

Let <math>f(x)=\sqrt{x}-1</math> and <math>g(x)=(f\circ f)(x)</math>. Find the derivative of <math>g(x)</math>.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 04 (b)/Solution 1

2014-11-15T01:18:21Z

ThomasWong: Created page with "Looking at the outside function, we want to find <math>\sqrt{x}-1\geq 0</math> for <math>\sqrt{\sqrt{x}-1}</math> to be defined. So we want <math>\sqrt{x} \geq 1</math>. Look..."

Looking at the outside function, we want to find <math>\sqrt{x}-1\geq 0</math> for <math>\sqrt{\sqrt{x}-1}</math> to be defined.

So we want <math>\sqrt{x} \geq 1</math>. Looking at the inside function, we want <math>\sqrt{x} \geq 0</math> so <math>x \geq 0</math>.

Combining the two, we get:<math>x\geq 0</math> and <math>x\geq 1</math>.

Hence the domain of <math>g(x)</math> is <math>\{x \mathbb{R} \mid x\geq 1 \}</math> or <math>x \in [1,\infty)</math>.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 04 (b)/Statement

2014-11-15T01:17:58Z

ThomasWong:

Let <math>f(x)=\sqrt{x}-1</math> and <math>g(x)=(f\circ f)(x)</math>. What is the domain of <math>g(x)</math>?

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 04 (b)/Statement

2014-11-15T01:16:25Z

ThomasWong: Created page with "Let <math>f(x)=\sqrt{x}-1</math> and <math>g(x)=(f\circ f)(x)</math>. Find the derivative of <math>g(x)</math>."

Let <math>f(x)=\sqrt{x}-1</math> and <math>g(x)=(f\circ f)(x)</math>. Find the derivative of <math>g(x)</math>.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 04 (a)/Solution 1

2014-11-15T01:15:26Z

ThomasWong: Created page with "We simply substitute: <math>\begin{align} g(x) &= (f\circ f)(x)\\ &= f(f(x))\\ &= f(\sqrt{x}-1)\\ g(x)&= \sqrt{\sqrt{x}-1}-1.\end{align}</math>"

We simply substitute:

<math>\begin{align}
g(x) &= (f\circ f)(x)\\
&= f(f(x))\\
&= f(\sqrt{x}-1)\\
g(x)&= \sqrt{\sqrt{x}-1}-1.\end{align}</math>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 04 (a)/Statement

2014-11-15T01:14:35Z

ThomasWong: Created page with "Let <math>f(x)=\sqrt{x}-1</math> and <math>g(x)=(f\circ f)(x)</math>. Find a formula for <math>g(x)</math> in terms of <math>x</math> (in other words, your formula should not ..."

Let <math>f(x)=\sqrt{x}-1</math> and <math>g(x)=(f\circ f)(x)</math>. Find a formula for <math>g(x)</math> in terms of <math>x</math> (in other words, your formula should not use the symbol “<math>f</math>”).

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 03 (b)/Solution 1

2014-11-15T01:10:54Z

ThomasWong: Created page with "'''Parts b and c combined:''' (These are all the rules I can think of). <ul> <li>Addition rule: <math>\begin{align} f(x) &= \frac{1}{2}x^6 + \frac{1}{2..."

'''Parts b and c combined:''' (These are all the rules I can think of).

<ul>
<li>Addition rule:
<math>\begin{align}
f(x) &= \frac{1}{2}x^6 + \frac{1}{2}x^6 &
f'(x) &= 3x^5 + 3x^5\end{align}</math></li>
<li>Subtraction rule:
<math>\begin{align}
f(x) &= 2x^6 - x^6 &
f'(x) &= 12x^5 - 6x^5\end{align}</math></li>
<li>Product rule:
<math>\begin{align}
f(x) &= x^2 \cdot x^4 &
f'(x) &= 2x \cdot x^4 + 4x^3 \cdot x^2\end{align}</math></li>
<li>Quotient rule:
<math>\begin{align}
f(x) &= \frac{x^7}{x} &
f'(x) &= \frac{7x^6 \cdot x - x^7 \cdot 1}{x^2}\end{align}</math></li>
<li>Chain rule:
<math>\begin{align}
f(x) &= (x^3)^2 &
f'(x) &= 3x^2 \cdot 2(x^3)\end{align}</math></li>
<li>Log-diff:
<math>\begin{align}
\ln(f(x)) &= 6 \ln(x)&
f'(x) &= \frac{6}{x}\cdot x^6\end{align}</math></li>
<li>Exp-chain rule:
<math>\begin{align}
f(x) &= \exp[6\ln(x)]&
f'(x) &= \frac{6}{x} \cdot \exp[6\ln(x)]\end{align}</math></li>
<li>Limit definition of derivative:
<math>\begin{align}
f'(x) &= \lim_{h\rightarrow 0} \frac{(x+h)^6 - x^6}{h} & f'(x) &= \lim_{h\rightarrow 0} \frac{6x^5h + 15x^4h^2 + \ldots + h^6}{h}\\
f'(x) &= \lim_{a\rightarrow x} \frac{a^6 - x^6}{a-x} & f'(x) &= \lim_{a\rightarrow x} \frac{(a-x)(a^5 + a^4x + \ldots x^5)}{(a-x)}\end{align}</math></li></ul>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 03 (b)/Statement

2014-11-15T01:09:20Z

ThomasWong: Created page with "Rewrite <math>f(x)</math> in such a way that you can differentiate it using a different rule than the Power Rule. Name the rule of differentiation which is applicable, and the..."

Rewrite <math>f(x)</math> in such a way that you can differentiate it using a different rule than the Power Rule. Name the rule of differentiation which is applicable, and then confirm that the derivative calculated using this alternative method is the same as in part (a).

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 03 (a)/Solution 1

2014-11-15T01:08:17Z

ThomasWong: Created page with "Using the power rule with <math>n=6</math>. <math>\begin{align} f'(x)&= 6x^5.\end{align}</math>"

Using the power rule with <math>n=6</math>.

<math>\begin{align}
f'(x)&= 6x^5.\end{align}</math>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 03 (a)/Statement

2014-11-15T01:07:32Z

ThomasWong: Created page with "What is the derivative of <math>f(x)</math>?"

What is the derivative of <math>f(x)</math>?

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 02/Solution 1

2014-11-15T01:06:03Z

ThomasWong: Created page with "We first recognise that parallel means that the slopes must be equal. Rearranging the equation, we get: <math>\begin{align} x-2y &= 2\\ x-2 &= 2y\\ y&= \frac{x-2}{2}\\ y &= \..."

We first recognise that parallel means that the slopes must be equal. Rearranging the equation, we get:

<math>\begin{align}
x-2y &= 2\\
x-2 &= 2y\\
y&= \frac{x-2}{2}\\
y &= \frac{1}{2} x - 1.\end{align}</math>

So that means we are looking for points along the curve with slope equal to <math>\frac{1}{2}</math>. To get those points, we need the derivative of the curve:

<math>\begin{align}
{\frac{{\rm d}}{{\rm d}x}} (y) &= {\frac{{\rm d}}{{\rm d}x}} \left(\frac{x-4}{x+4}\right)\end{align}</math>

Using the quotient rule, we get:

<math>\begin{align}
u(x) &= x-4 & u'(x) = 1\\
v(x) &= x+4 & v'(x) = 1\end{align}</math>

So that means:

<math>\begin{align}
{\frac{{\rm d} {y}}{{\rm d}x}} &= \frac{v(x)u'(x) - u(x)v'(x)}{\left(v(x)\right)^2}\\
&= \frac{(x+4)\cdot (1) - (x-4) \cdot(1)}{\left(x+4\right)^2}\\
&= \frac{8}{\left(x+4\right)^2}\end{align}</math>

We are interested in points where the derivative is <math>\frac{1}{2}</math>

<math>\begin{align}
\frac{1}{2} &= \frac{8}{\left(x+4\right)^2}\\
\left(x+4\right)^2 &= 16\\
x+4 &= \pm 4\\
x &= 0 , -8\end{align}</math>

This means we have to tackle two potential <math>x</math> values. At <math>x=0</math>, we have:

<math>\begin{align}
y &= \frac{0-4}{0+4} = -1\end{align}</math>

At <math>x=-8</math>, we have:

<math>\begin{align}
y &= \frac{-8-4}{-8+4} = \frac{-12}{-4} = 3\end{align}</math>

So that means the tangent lines are are looking for are: for <math>x=0</math>.

<math>\begin{align}
y &= \frac{1}{2}(x-0) +(-1)\\
&= \frac{1}{2}x - 1\\\end{align}</math>

and for <math>x=-8</math>

<math>\begin{align}
y &= \frac{1}{2}(x-(-8)) + 3\\
&= \frac{1}{2} (x+8) + 3\\
&= \frac{1}{2} x + 7\\\end{align}</math>

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 01 (d)/Solution 1

2014-11-15T00:56:33Z

ThomasWong: Created page with "'''False:''' If <math>f(x)</math> is differentiable at <math>x=2</math>, then by the definition of derivative at a point, we have <math>\displaystyle f'(2) = \lim..."

'''False:''' If <math>f(x)</math> is differentiable at <math>x=2</math>, then by the definition of derivative at a point, we have <math>\displaystyle f'(2) = \lim_{h\rightarrow0}\frac{f(2+h)-f(2)}{h}.</math> So to paraphrase the statement, we are asking whether <math>f'(2) = 2</math>. Clearly, this does not have to be the case. Consider the example <math>f(x)=1</math>. We have <math>f(x)</math> is differentiable at <math>x=2</math> (and everywhere else too). But <math>f'(2) = 0</math>.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 01 (c)/Solution 1

2014-11-15T00:55:35Z

ThomasWong: Created page with "'''False:''' We can use the same function as the previous part (if we want) to show that we can have every point leading to <math>0</math> to be defined at <math>..."

'''False:''' We can use the same function as the previous part (if we want) to show that we can have every point leading to <math>0</math> to be defined at <math>2</math> but <math>f(0)</math> to be undefined. Here's another counter-example: <math>f(x) = \frac{2x}{x}</math>. We have <math>f(x) = 2</math> for all <math>x</math> except for <math>0</math> and <math>f(0)</math> to be undefined.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 01 (b)/Solution 1

2014-11-15T00:53:36Z

ThomasWong: Created page with "'''False:''' The easiest way to see this is to come up with a function where the limit exists but the function is not defined there. For example <math>f(x) = \fra..."

'''False:''' The easiest way to see this is to come up with a function where the limit exists but the function is not defined there. For example <math>f(x) = \frac{4(x-1)}{x-1}</math>. The limit exists and is equal to <math>4</math> but <math>1</math> is not in the domain, so <math>f(1)</math> is not defined.

Science:Math Exam Resources/Courses/MATH110/December 2013/Question 01 (a)

2014-11-15T00:52:23Z

ThomasWong:

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