Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 04 (b)/Solution 1

Observe that $f$ can be written as a composition of two functions $g(x)=\sec x$ and $h(x)={\frac {1}{x}}$;

$$f(x)=\sec \left({\frac {1}{x}}\right)=g(h(x))$$

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By Hint 2 and the power rule, we have $g'(x)=(\sec x)'=\sec x\tan x$ and $h'(x)=(x^{-1})'=-x^{-2}$.

Then, we apply the chain rule to get

$$f'(x)=g'(h(x))h'(x)=\sec \left({\frac {1}{x}}\right)\tan \left({\frac {1}{x}}\right)\cdot \left(-{\frac {1}{x^{2}}}\right)$$

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Finally, plug $x={\frac {4}{\pi }}$ into $f'(x)$, we get

$$f'\left({\frac {4}{\pi }}\right)=-\sec \left({\frac {\pi }{4}}\right)\tan \left({\frac {\pi }{4}}\right)\cdot \left({\frac {\pi }{4}}\right)^{2}=-{\frac {{\sqrt {2}}\pi ^{2}}{16}}$$

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In the last equality, $\sec \left({\frac {\pi }{4}}\right)={\sqrt {2}}$ and $\tan \left({\frac {\pi }{4}}\right)=1$ are used.

Answer: $f'\left({\frac {4}{\pi }}\right)=\color {blue}{-{\frac {{\sqrt {2}}\pi ^{2}}{16}}}$