Question

Find the derivative of ${\sin }^{n}x,$ where $n$ is an integer.

Answer 1

Method $1:$
Calculating derivative for $p^n$
$f\left(x\right)=p^n$
where $p$ be a function in $x$.
For $n=1$
$f\text{'}\left(x\right)=p\text{'}$
For $n=2$
$f\text{'}\left(x\right)=\left(p^2\right)\text{'}=(p\cdot p)\text{'}=pp\text{'}+p\text{'}p=2p{p}^{\prime }$
For $n=3$
$f^{\prime }\left(x\right)=(p^3{)}^{\prime }=(p\cdot p^2{)}^{\prime }$
$\Rightarrow f^{\prime }\left(x\right)=p(p^2{)}^{\prime }+p^{\prime }{p}^2=2p^2{p}^{\prime }+p^2{p}^{\prime }$
$\Rightarrow f^{\prime }\left(x\right)=3p^2{p}^{\prime }$
Similarly, for $n=n$
$f^{\prime }\left(x\right)=(p^n{)}^{\prime }$
$\Rightarrow f^{\prime }\left(x\right)=n{p}^{n-1}{p}^{\prime }$

Required derivative
Let $g\left(x\right)={\sin }^{n}x$
Differentiating with respect to $x$
$\Rightarrow g^{\prime }\left(x\right)=({\sin }^{n}x{)}^{\prime }=((\sin x{)}^n{)}^{\prime }$
From the above formula, we get
$g^{\prime }\left(x\right)=n\left(\sin x{)}^{n-1}\right(\sin x{)}^{\prime }$
$\therefore g^{\prime }\left(x\right)=n{\sin }^{n-1}x\cos x$