Answer to Question #223105 in Calculus for Sama

Question

Find f'(x) if

f(x)= √[In(e^2x + e^-2x)]

f(x) = e^-2x sin2x

Answer 1

(1) Given, $f(x)= \sqrt{ln(e^{2x}+e^{-2x})}$

Differentiating with respect to x,

$f'=\frac{df}{dx}=\frac{1}{2\sqrt{ln(e^{2x}+e^{-2x})}}\frac{(2e^{2x}-2e^{-2x})}{(e^{2x}+e^{-2x})} = \frac{(e^{2x}-e^{-2x})}{(e^{2x}+e^{-2x})\sqrt{ln(e^{2x}+e^{-2x})}}$

(2) Given, $f(x) = e^{-2x} sin2x$

Differentiating both sides with respect to x,

$f'(x) = \frac{df}{dx}=2e^{-2x}cos2x -2e^{-2x}sin2x = 2[e^{-2x}cos2x-e^{-2x}sin2x]$