Science:Math Exam Resources/Courses/MATH104/December 2014/Question 02

{{#incat:MER QGQ flag|{{#incat:MER QGH flag|{{#incat:MER QGS flag|}}}}}}

MATH104 December 2014

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q1 (k)  •  Q1 (l)  •  Q1 (m)  •  Q1 (n)  •  Q2  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q5 (d)  •  Q5 (e)  •  Q5 (f)  •  Q6  •  

   Other MATH104 Exams

  •  December 2014  •  December 2011  •  December 2010  •  December 2012  •  December 2013  •  December 2015  •  December 2016  •  

Question 02
Let $f(x)={\frac {x^{2}}{x^{2}+1}}$. Use the definition of derivative to find $f'(1)$. No marks will be given for the use of any differentiation rules.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
There are two equivalent limit definitions of the derivative at the point $a$: $f'(a)=\lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}$ and $f'(a)=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}$ given that these limits exist. You can use either definition. See Solution 1 for the former approach, and Solution 2 for the latter.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

Here we first calculate the derivative for any value $x$ and only then plug in the value $a=1$. That is, we first calculate $f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}$ We use this to evaluate the derivative of the given function: ${\begin{aligned}f'(x)&=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {{\frac {(x+h)^{2}}{(x+h)^{2}+1}}-{\frac {x^{2}}{x^{2}+1}}}{h}}\\&=\lim _{h\rightarrow 0}{\frac {(x+h)^{2}(x^{2}+1)-x^{2}((x+h)^{2}+1)}{h((x+h)^{2}+1)(x^{2}+1)}}\\&=\lim _{h\rightarrow 0}{\frac {(x+h)^{2}x^{2}+(x+h)^{2}-x^{2}(x+h)^{2}-x^{2}}{h((x+h)^{2}+1)(x^{2}+1)}}\\&=\lim _{h\rightarrow 0}{\frac {(x+h)^{2}-x^{2}}{h((x+h)^{2}+1)(x^{2}+1)}}\\&=\lim _{h\rightarrow 0}{\frac {x^{2}+2xh+h^{2}-x^{2}}{h((x+h)^{2}+1)(x^{2}+1)}}\\&=\lim _{h\rightarrow 0}{\frac {h(2x+h)}{h((x+h)^{2}+1)(x^{2}+1)}}\\&=\lim _{h\rightarrow 0}{\frac {(2x+h)}{((x+h)^{2}+1)(x^{2}+1)}}\\&={\frac {2x}{(x^{2}+1)^{2}}}\end{aligned}}$ Lastly, plugging in we obtain $f'(1)={\frac {2(1)}{(1^{2}+1)^{2}}}={\frac {1}{2}}$.

Solution 2

In this solution we use the formula $f'(a)=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}$ We could calculate $f'(a)$ for any value of $a$ and then plug in $a=1$, as it was done in Solution 1, but to show an alternative we plug in $a=1$ directly. Following the above definition of the derivative and the definition of the function we obtain: ${\begin{aligned}f'(1)&=\lim _{x\to 1}{\frac {f(x)-f(1)}{x-1}}\\&=\lim _{x\to 1}{\frac {{\frac {x^{2}}{x^{2}+1}}-{\frac {1^{2}}{1^{2}+1}}}{x-1}}\\&=\lim _{x\to 1}{\frac {{\frac {2x^{2}}{2(x^{2}+1)}}-{\frac {x^{2}+1}{2(x^{2}+1)}}}{x-1}}\\&=\lim _{x\to 1}{\frac {2x^{2}-(x^{2}+1)}{2(x-1)(x^{2}+1)}}\\&=\lim _{x\to 1}{\frac {x^{2}-1}{2(x-1)(x^{2}+1)}}\\&=\lim _{x\to 1}{\frac {(x+1)(x-1)}{2(x-1)(x^{2}+1)}}\\&=\lim _{x\to 1}{\frac {x+1}{2(x^{2}+1)}}\\&={\frac {1+1}{2(1^{2}+1)}}\\&={\frac {1}{2}}\end{aligned}}$

{{#incat:MER CT flag||

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER QGT flag, MER Tag Limit definition of the derivative, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag

}}