Question

Differentiate from first principle:

$\left(ix\right)x\sin x$

Answer 1

Given:

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

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:


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

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

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


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

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

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

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

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

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