MATH 180 December 2017
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Question 07

Let ${\textstyle f(x)={\frac {1}{g(x)}}}$, where ${\textstyle g(x)}$ is a positive differentiable function. Show that $f'(x)={\frac {g'(x)}{g(x)^{2}}}$ using the limit definition of derivative. No credit will be given for using any other method.

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

Recall that ${\textstyle f'(x)=\lim \limits _{h\to 0}{\dfrac {f(x+h)f(x)}{h}}}$.

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Solution

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By the limit definition, we have
$f'(x)=\lim \limits _{h\to 0}{\dfrac {f(x+h)f(x)}{h}}.$
Since $f$ is given by $f(x)={\dfrac {1}{g(x)}}$, it becomes
$\lim \limits _{h\to 0}{\dfrac {f(x+h)f(x)}{h}}=\lim \limits _{h\to 0}{\dfrac {{\frac {1}{g(x+h)}}{\frac {1}{g(x)}}}{h}}=\lim \limits _{h\to 0}{\dfrac {1}{h}}\left({\dfrac {1}{g(x+h)}}{\dfrac {1}{g(x)}}\right).$
Making a common denominator, the expression inside of bracket becomes
${\frac {1}{g(x+h)}}{\frac {1}{g(x)}}={\frac {g(x)g(x+h)}{g(x)g(x+h)}}.$
Plugging this back, we have
$\lim \limits _{h\to 0}{\frac {1}{h}}\left({\frac {1}{g(x+h)}}{\frac {1}{g(x)}}\right)=\lim \limits _{h\to 0}{\frac {1}{h}}\left({\frac {g(x)g(x+h)}{g(x)g(x+h)}}\right)=\lim \limits _{h\to 0}{\frac {g(x)g(x+h)}{h\cdot g(x)g(x+h)}}$ .
We can rewrite the fraction as
${\frac {g(x)g(x+h)}{h\cdot g(x)g(x+h)}}={\frac {1}{g(x)g(x+h)}}\cdot \left({\frac {g(x+h)g(x)}{h}}\right).$
Since $\lim _{h\to 0}g(x)=g(x)g(x+h)=g(x)^{2}$ and $g'(x)=\lim \limits _{h\to 0}{\frac {g(x+h)g(x)}{h}}$, using the product and quotient laws of the limits, we get
${\begin{aligned}\lim \limits _{h\to 0}{\frac {1}{g(x)g(x+h)}}\cdot \left({\frac {g(x+h)g(x)}{h}}\right)&=\lim \limits _{h\to 0}{\frac {1}{g(x)g(x+h)}}\cdot \left(\lim \limits _{h\to 0}{\frac {g(x+h)g(x)}{h}}\right)\\&={\frac {1}{\lim \limits _{h\to 0}g(x)g(x+h)}}\cdot \left(\lim \limits _{h\to 0}{\frac {g(x+h)g(x)}{h}}\right)\\&={\frac {1}{g(x)^{2}}}\cdot {(g'(x))}={\frac {g'(x)}{g(x)^{2}}}.\end{aligned}}$
Therefore, we get
$f'(x)={\frac {g'(x)}{g(x)^{2}}}.$
