Question

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

Answer 1

Since, $y=e^{x \sin x^{3}}+(\tan x)^{x}$, then
$y=u+v$, where $u=e^{x \sin x^{3}}$ and $v=(\tan x)^{x}$
$\Rightarrow \frac{d y}{d x}=\left(\frac{d u}{d x}+\frac{d v}{d x}\right) ....$(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{d u}{d x} =e^{x \sin x^{3}} \cdot\left(3 x^{3} \cos x^{3}+\sin x^{3}\right)... $(ii)
and $ \frac{1}{v} \cdot \frac{d v}{d x} =\frac{x \cdot \sec ^{2} x}{\tan x}+\log (\tan x) $
$\frac{d v}{d x} =(\tan x)^{x}[2 x \cdot \operatorname{cosec}(2 x)+\log (\tan x)] \ldots $ (iii)
From Eqs. (i), (ii) and (iii), wet get
$\frac{d y}{d x}=e^{x \sin x^{3}}\left(3 x^{3} \cdot \cos x^{3}+\sin x^{3}\right)+(\tan x)^{x} $
$[2 x \operatorname{cosec} 2 x+\log (\tan x)]$