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.

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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
$f'(x)={\frac {g'(x)}{g(x)^{2}}}.$


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