Question

Differentiate from first principle:

$\left(ii\right)\frac{\sin x}{x}$

Answer 1

Given: $f\left(x\right)=\frac{\sin x}{x}$

The derivative of a function $f\left(x\right)$ is defined as:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

Putting $f\left(x\right)$ in above expression, we get:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{\sin (x+h)}{x+h}-\frac{\sin x}{x}}{h}$


$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{x\sin (x+h)-(x+h)\sin x}{hx(x+h)}$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{x(\sin x\cos h+\cos x\sin h)-x\sin x-h\sin x}{hx(x+h)}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{x\sin x(\cos h-1)+x\cos x\sin h-h\sin x}{hx(x+h)}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{-2x\sin x{\sin }^2\frac{h}{2}}{hx(x+h)}+\underset{h\to 0}{lim}\frac{x\cos x\sin h}{hx(x+h)}+\underset{h\to 0}{lim}\frac{-h\sin x}{hx(x+h)}$

$\Rightarrow f^{\prime }\left(x\right)=-2\sin x\underset{h\to 0}{lim}\frac{1}{(x+h)}.\frac{{\sin }^2\frac{h}{2}}{{\left(\frac{h}{2}\right)}^2}.\frac{{\left(\frac{h}{2}\right)}^2}{h}+\underset{h\to 0}{lim}\frac{\cos x}{(x+h)}.\frac{\sin h}{h}+\underset{h\to 0}{lim}\frac{-\sin x}{x(x+h)}$

$\Rightarrow f^{\prime }\left(x\right)=-2\sin x\underset{x\to 0}{lim}\frac{1}{(x+h)}.\frac{h}{4}+\frac{\cos x}{x+0}-\frac{\sin x}{x(x+0)}[\because \underset{h\to 0}{lim}\frac{\sin h}{h}=1]$

$\Rightarrow f^{\prime }\left(x\right)=\frac{\cos x}{x}-\frac{\sin x}{x^2}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{x\cos x-\sin x}{x^2}$