Answer to Question #226792 in Calculus for Amorah

Question

(Cos(2x))^1/(x^2) please help with the derivative. Thanks

Answer 1

$$y=(\cos(2x))^{1/x^2}$$

$$\ln y=\dfrac{\ln(\cos(2x))}{x^2}$$

Differentiate both sides with respect to $x$

$$\dfrac{y'}{y}=\dfrac{-2x^2\tan(2x)-2x\ln(\cos(2x))}{x^4}$$

$$y'=-\dfrac{2}{x^3}\big(x\tan(2x)+\ln(\cos(2x))\big)\big(\cos(2x)\big)^{1/x^2}$$