Question

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

Answer 1

Method $1:$ Calculating derivative for $p^n$
Let $f\left(x\right)=p^n$ 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 $n=3$
$f\text{'}\left(x\right)=\left(p^3\right)\text{'}=(p\cdot p^2)\text{'}=p\left(2p{p}^{\prime }\right)+p\text{'}{p}^2$
$\Rightarrow f^{\prime }\left(x\right)=2p^2{p}^{\prime }+p^2{p}^{\prime }$
$\Rightarrow f^{\prime }\left(x\right)=3p^{2}p\text{'}$
Similarly, for $n=n,$
$f\text{'}\left(x\right)=\left(p^n\right)\text{'}$
$\Rightarrow f^{\prime }\left(x\right)=n{p}^{n-1}p\text{'}$

Calculating derivative of ${\sin }^{n}x$
Let $g\left(x\right)={\sin }^{n}x$
Differentiating with respect to $x$
$\Rightarrow g\text{'}\left(x\right)=\left({\sin }^{n}x\right)\text{'}=(\sin x{)}^n{)}^{\prime }$
From above formula
$g\text{'}\left(x\right)=n\left(\sin x{)}^{n-1}\right(\sin x{)}^{\prime }$
$\Rightarrow g\text{'}\left(x\right)=n\left(\sin x{)}^{n-1}\right(\cos x)$
$\Rightarrow g\text{'}\left(x\right)=n{\sin }^{n-1}x\cos x$
Required derivative
Let $p\left(x\right)=\frac{x}{{\sin }^{n}x}$
Differentiating with respect to $x$
$\Rightarrow p\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{x}{{\sin }^{n}x}\right)$
$\Rightarrow p^{\prime }\left(x\right)=\frac{\left({\sin }^{n}x\right)\frac{d}{dx}\left(x\right)-\left(x\right)\frac{d}{dx}\left({\sin }^{n}x\right)}{({\sin }^{n}x{)}^2}$
$\Rightarrow p^{\prime }\left(x\right)=\frac{1\left({\sin }^{n}x\right)-x\left(n{\sin }^{n-1}x\cos x\right)}{({\sin }^{n}x{)}^2}$
$\Rightarrow p^{\prime }\left(x\right)=\frac{{\sin }^{n-1}x(\sin x-n\cdot x\cos x)}{{\sin }^{2n}x}$
$\Rightarrow p^{\prime }\left(x\right)=\frac{\sin x-n\cdot x\cos x}{{\sin }^{n+1}x}$
$\Rightarrow p^{\prime }\left(x\right)=\frac{\sin x-n.x\cos x}{{\sin }^{n+1}x}$

Method $2:$
Let $p\left(x\right)=\frac{x}{{\sin }^{n}x}$
Differentiating with respect to $x$
$\Rightarrow p^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{x}{{\sin }^{n}x}\right)$
$\Rightarrow p^{\prime }\left(x\right)=\frac{\left({\sin }^{n}x\right)\frac{d}{dx}\left(x\right)-\left(x\right)\frac{d}{dx}\left({\sin }^{n}x\right)}{({\sin }^{n}x{)}^2}$
Using Chain Rule, we get
$\Rightarrow p^{\prime }\left(x\right)=\frac{1\left({\sin }^{n}x\right)-x\left(n{\sin }^{n-1}x\cos x\right)}{({\sin }^{n}x{)}^2}$
$\Rightarrow p^{\prime }\left(x\right)=\frac{{\sin }^{n-1}x(\sin x-n\cdot x\cdot \cos x)}{{\sin }^{2n}x}$
$\Rightarrow p^{\prime }\left(x\right)=\frac{\sin x-nx\cos x}{{\sin }^{2n}x\cdot {\sin }^{-(n-1)x}}$
$\Rightarrow p^{\prime }\left(x\right)=\frac{\sin x-nx\cos x}{{\sin }^{2n-n+1}x}$
$\therefore p^{\prime }\left(x\right)=\frac{\sin x-nx\cos x}{{\sin }^{n+1}x}$