Question

Differentiate from first principle:

$\left(iii\right)\frac{\cos x}{x}$

Answer 1

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


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:

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

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

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

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

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


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

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

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

Therefore, the derivative of $\frac{\cos x}{x}\text{is}$
$\frac{-x\sin x-\cos x}{x^2}.$