Question

Determine the values of x for which the derivative of the function f(x) = cos(x+(1/x)) is 0

Answer 1

\[f(x) = \cos(x+\frac{1}{x})\]

Using chain rule
\[f'(x) = -(1-\frac{1}{x^2})\sin(x+\frac{1}{x})\] so u have to solve the equation \[ (1-\frac{1}{x^2})\sin(x+\frac{1}{x})=0\]

Answer 2

note that the equation\[\sin x=0\]has the answers in the form \(x=n\pi\) where \(n \in \mathbb{Z}\)

Answer 3

great then i equate x to x+ (1/x) correct?

Answer 4

Yeah...
if u wanna check .... the final answer is

\[x=\pm1\]and \[x=\frac{n\pi\pm\sqrt{n^2\pi^2-4}}{2} .........n \in \mathbb{Z}-\left\{ 0 \right\} \]