Calculus problem on functions for finding $a$ Question:
Given a function $$f(x)=(60073-x^{10})^{1/10}$$ and $$f'(2)=\frac{1}{f'(a)},$$ where $a$ is a positive integer, find $a$.
My approach:
I can proceed with the problem by differentiating the equation and finding $f'(x)$ and then put $2$ and $a$. But I was wondering if there is any shorter method to solve these problems since these problems was generally given in competitive examinations.
Any help would be appreciated.
 A: There's a few keys to this question. The first is that $a$ is a positive integer. Also notice that $f(2) = (60073-2^{10})^{1/10} = (3^{10})^{1/10}=3$, and that if $n$ is an integer greater than $3$, $f(n)$ is imaginary. This already hints at what the answer is.
Now it is possible differentiate the function straightaway to find $f(x)$. I think taking the $\log$ of both sides first is easier. Then we have
$$\log f(x) = \frac{1}{10} \log (60073-x^{10}) $$
and differentiating with respect to $x$,
$$ \frac{f'(x)}{f(x)} = \frac{-x^9}{60073-x^{10}} = \frac{-x^9}{[f(x)]^{10}}$$
So we have
$$f'(x) = \frac{-x^9}{[f(x)]^9}$$
Now we use the given information,
$$f'(2) = \frac{-2^9 }{f(2)^9} = -\frac{2^9}{3^9} $$
Then we need to solve for $a$ with
$$ \frac{1}{f'(a)} = -\frac{f(a)^9}{a^9} = -\frac{2^9}{3^9} $$
Then it's clear $a=3$ is an integer solution.
A: Hint:  $\displaystyle f'(a)=\frac{1}{f'(2)}$ is equivalent to $\displaystyle  - \left(\frac{a}{\sqrt[10]{60073 - a^{10}}}\right)^9 = - \frac{19683}{512}= -\left(\frac{3}{2}\right)^{9}\,$.
Then $\displaystyle \frac{a}{\sqrt[10]{60073 - a^{10}}} = \frac{3}{2} \iff 2^{10} a^{10} = 3^{10}(60073 - a^{10}) \iff a^{10}=59049  \iff a=3$.
