Question

$y=e^{\{{\sin }^{2}x+{\sin }^{4}x+{\sin }^{6}x+...\}}$, then $\frac{dy}{dx}=$

Answer 1

$\frac{a}{1-r}$, where $a={\sin }^{2}x$
$r={\sin }^{2}x$
$=\frac{{\sin }^{2}x}{1-{\sin }^{2}x}={\tan }^{2}x$

$y=e^{{\tan }^{2}x}$

Differentiating with respect to x,

$\frac{dy}{dx}=2e^{{\tan }^{2}x}\tan x{\sec }^{2}x$