Question

Let $y=e^{x\sin x^3}+{\left(\tan x\right)}^x$ find $\frac{dy}{dx}$ (IIT-JEE, 1981)

Answer 1

We are given that $y=e^{x\sin x^3}+{\left(\tan x\right)}^x$

$u+v$

$u=e^{x\sin x^3}$ and $v={\left(\tan x\right)}^x$

now $\frac{du}{dx}=e^{x\sin x^3}\frac{d}{dx}\left(x\sin x^3\right)$

$=e^{x\sin x^3}[3x^3\times \cos x^3+\sin x^3]$

$v={\left(\tan x\right)}^x$

Differentiating w.r.t. $x$, we get $\frac{1}{v}\frac{dv}{dx}=x\frac{1}{\tan x}{\sec }^{2}x+\log \tan x$

$\therefore \frac{dv}{dx}={\left(\tan x\right)}^x(\frac{2x}{\sin 2x}+\log \tan x)$

Hence, $\frac{dy}{dx}=e^{x\sin x^3}(\sin x^3+3x^3\cos x^3)+{\left(tanx\right)}^x(\frac{2x}{sin2x}+\log \tan x)$