Question

Differentiate from first principle:

$\left(iv\right)x^2\sin x$

Answer 1

Given: $f\left(x\right)=x^2\sin x$

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 above expression, we get:

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


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

$=\underset{h\to 0}{lim}\frac{x^2\sin x(\cos h-1)}{h}+\underset{h\to 0}{lim}\frac{x^2\cos x\sin h}{h}+\underset{h\to 0}{lim}\frac{(h^2+2xh)(\sin x\cos h+\cos x\sin h)}{h}$

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

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

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

$=x^2\cos x+2x\sin x$