Question

Differentiate from first principle:

$\left(x\right)x\cos x$

Answer 1

Givne:

$f\left(x\right)=x\cos x$

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 the above expression, we get:


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

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


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

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

$[\because \underset{h\to 0}{lim}\frac{\sin h}{h}=1]$

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

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

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