MATH104 December 2011
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Question 01 (c)

Find $f'(x)$ where
$f(x)={\frac {(x^{2}+3\sin ^{2}(x))(e^{x^{2}})}{x^{4}+7}}$
Do NOT simplify your answer.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1

Which differentiation rules do you need to solve this question? You'll need more than one.

Hint 2

Try using the quotient rule first, and then the chain rule as needed.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
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Solution 1

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One method of solving this problem is to use the quotient rule:
 $f'(x)={\frac {\left({\frac {d}{dx}}{\Big [}(x^{2}+3\sin ^{2}x)(e^{x^{2}}){\Big ]}\right)(x^{4}+7)(x^{2}+3\sin ^{2}x)(e^{x^{2}}){\bigl (}{\frac {d}{dx}}(x^{4}+7){\bigr )}}{(x^{4}+7)^{2}}}$
For the first derivative we need the product rule (and then the chain rule):
 ${\begin{aligned}{\frac {d}{dx}}{\Big [}(x^{2}+3\sin ^{2}x)(e^{x^{2}}){\Big ]}&={\bigl (}{\frac {d}{dx}}(x^{2}+3\sin ^{2}x){\bigr )}e^{x^{2}}+(x^{2}+3\sin ^{2}x){\bigl (}{\frac {d}{dx}}e^{x^{2}}{\bigr )}\\&=(2x+6\sin x\cos x)e^{x^{2}}+(x^{2}+3\sin ^{2}x)(2xe^{x^{2}})\end{aligned}}$
Then, by substitution
 ${\begin{aligned}f'(x)&={\frac {\left((2x+6\sin x\cos x)e^{x^{2}}+(x^{2}+3\sin ^{2}x)(2xe^{x^{2}})\right)(x^{4}+7)(x^{2}+3\sin ^{2}x)e^{x^{2}}4x^{3}}{(x^{4}+7)^{2}}}\\\end{aligned}}$
Luckily, no further simplification is required for this question.

Solution 2

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Another method of solving this problem is to use the product rule. First rewrite the denominator of the given function using exponent notation
 ${\begin{aligned}f(x)&={\frac {(x^{2}+3\sin ^{2}x)(e^{x^{2}})}{x^{4}+7}}\\&=(x^{2}+3\sin ^{2}x)(e^{x^{2}})(x^{4}+7)^{1}\end{aligned}}$
This is a product of three functions. The product rule can be generalized from its twofunction version to:
 ${\frac {d}{dx}}(f\cdot g\cdot h)={\frac {df}{dx}}\cdot g\cdot h+f\cdot {\frac {dg}{dx}}\cdot h+f\cdot g\cdot {\frac {dh}{dx}}$
So taking the derivative, we obtain
 ${\begin{aligned}f'(x)&={\frac {d}{dx}}{\Big (}(x^{2}+3\sin ^{2}x)(e^{x^{2}})(x^{4}+7)^{1}{\Big )}\\&={\Big (}{\frac {d}{dx}}(x^{2}+3\sin ^{2}x){\Big )}(e^{x^{2}})(x^{4}+7)^{1}\\&\quad +(x^{2}+3\sin ^{2}x){\Big (}{\frac {d}{dx}}(e^{x^{2}}){\Big )}(x^{4}+7)^{1}\\&\quad +(x^{2}+3\sin ^{2}x)(e^{x^{2}}){\Big (}{\frac {d}{dx}}(x^{4}+7)^{1}{\Big )}\\&={\Big (}2x+6\sin x\cos x{\Big )}(e^{x^{2}})(x^{4}+7)^{1}\\&\quad +(x^{2}+3\sin ^{2}x){\Big (}2xe^{x^{2}}{\Big )}(x^{4}+7)^{1}\\&\quad +(x^{2}+3\sin ^{2}x)(e^{x^{2}}){\Big (}4x^{3}(x^{4}+7)^{2}{\Big )}\end{aligned}}$
No further simplification is required, for this question.

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