Question
Differentiate the following w.r.t. x :

$y =x^{\frac{4}{3}}+e^x-\sin x$

Answer

$
y=x^{\frac{4}{3}}+ e ^x-\sin x
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{ d y}{ d x} & =\frac{ d }{ d x}\left(x^{\frac{4}{3}}+ e ^x-\sin x\right) \\
\frac{ d y}{ d x} & =\frac{ d }{ d x}\left(x^{\frac{4}{3}}\right)+\frac{ d }{ d x}\left( e ^x\right)-\frac{ d }{ d x}(\sin x) \\
& =\frac{4}{3} x^{\frac{4}{3}-1}+ e ^x-\cos x \\
& =\frac{4}{3} x^{\frac{1}{3}}+ e ^x-\cos x
\end{aligned}
$

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