Question

Let $y=e^{x\sin x^3}+(\tan x{)}^x$, find $\frac{dy}{dx}$.

Answer 1

Since, $y=e^{x\sin x^3}+(\tan x{)}^x$, then
$y=u+v$, where $u=e^{x\sin x^3}$ and $v=(\tan x{)}^x$
$\Rightarrow \frac{dy}{dx}=\left(\frac{du}{dx}+\frac{dv}{dx}\right)....$(i)
Here, $u=e^{x\sin x^3}$ and $\log v=x\log (\tan x)$
On differentiating both sides w.r.t. $x$, we get
$\frac{du}{dx}=e^{x\sin x^3}\cdot \left(3x^3\cos x^3+\sin x^3\right)...$(ii)
and $\frac{1}{v}\cdot \frac{dv}{dx}=\frac{x\cdot {\sec }^{2}x}{\tan x}+\log (\tan x)$
$\frac{dv}{dx}=(\tan x{)}^x[2x\cdot \mathrm{cosec}(2x)+\log (\tan x)]\dots$ (iii)
From Eqs. (i), (ii) and (iii), wet get
$\frac{dy}{dx}=e^{x\sin x^3}\left(3x^3\cdot \cos x^3+\sin x^3\right)+(\tan x{)}^x$
$[2x\mathrm{cosec}2x+\log (\tan x)]$