Question

Differentiate $\frac{1}{3}{\tan }^{3}x-\tan x+x$ w.r.t $x$

Answer 1

Let $y=\frac{1}{3}{\tan }^{3}x-tanx+x$

Differentiating w.r.t $x$

$\frac{dy}{dx}=3.\frac{1}{3}{\tan }^{2}x.{\sec }^{2}x-{\sec }^{2}x+1$

$\frac{dy}{dx}={\tan }^{2}x{\sec }^{2}x-{\sec }^{2}x+1$

$\frac{dy}{dx}={\tan }^{2}x{\sec }^{2}x-({\sec }^{2}x-1)$

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

$\frac{dy}{dx}={\tan }^{2}x({\sec }^{2}x-1)$

$\frac{dy}{dx}={\tan }^{4}x$