Differentiate x^2 x from first principle.

Question

Differentiate $x^2 \sin x$ from first principle.

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Answer

We have to find derivative of $f(x)=x^2 \sin x$
Derivative of a function $f ( x )$ is given by $f ^{\prime}( x )=\underset{{h \rightarrow 0}}{\lim}=\frac{f(x+h)-f(x)}{h}$ where $h$ is a very small positive number
$\therefore$ Derivative of $f ( x )= x ^2 \sin x$ is given as $f ^{\prime}( x )\underset{{h \rightarrow 0}}{\lim}=\frac{f(x+h)-f(x)}{h}$
$\Rightarrow f ( x )\underset{{h \rightarrow 0}}{\lim} \frac{(x+h)^2 \sin (x+h)-x^2 \sin x}{h}$
$\text { Using }( a + b )^2= a ^2+2 ab + b ^2 \text {, we get }$
$\Rightarrow f ( x )\underset{{h \rightarrow 0}}{\lim} \frac{h^2 \sin (x+h)+x^2 \sin (x+h)+2 h x \sin (x+h)-x^2 \sin x}{h}$
Using the algebra of limits, we have
$\Rightarrow f ( x )\underset{{h \rightarrow 0}}{\lim} \frac{h^2 \sin (x+h)}{h}+\lim _{h \rightarrow 0} \frac{x^2 \sin (x+h)-x^2 \sin x}{h}+\lim _{h \rightarrow 0} \frac{2 h x \sin (x+h)}{h}$
$\Rightarrow f ( x )\underset{{h \rightarrow 0}}{\lim} h \sin ( x + h )+\lim _{h \rightarrow 0} \frac{x^2(\sin (x+h)-\sin x)}{h}+\lim _{h \rightarrow 0} 2 x \sin ( x + h )$
$\Rightarrow f ^{\prime}( x )=0 \times \sin ( x +0)+2 x \sin ( x +0)+ x ^2 \lim _{h \rightarrow 0} \frac{(\sin (x+h)-\sin x)}{h}$
$\Rightarrow f ^{\prime}( x )=2 x \sin + x ^2 \lim _{h \rightarrow 0} \frac{(\sin (x+h)-\sin x)}{h}$
Using the algebra of limits we have
$\therefore f^{\prime}(x)=2 x \sin x+x^2 \lim _{h \rightarrow 0} \frac{(\sin (x+h)-\sin x)}{h}$
We can't evaluate the limits at this stage only as on putting value it will take $\frac{0}{0}$ form.
So, we need to do little modifications.
$\text { Use: } \sin A-\sin B=2 \cos \left(\frac{(A-B)}{2}\right) \sin \left(\frac{(A-B)}{2}\right)$
$\therefore f^{\prime}(x)=2 x \sin x+x^2 \lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{2 x+h}{2}\right) \sin \left(\frac{h}{2}\right)}{h}$
$\Rightarrow f^{\prime}(x)=2 x \sin x+x^2 \lim _{h \rightarrow 0} \frac{\cos \left(x+\frac{h}{2}\right) \sin \left(\frac{A}{2}\right)}{\frac{A}{2}}$
Using the algebra of limits:
$\Rightarrow f^{\prime}(x)=2 x \sin x+x^2 \lim _{h \rightarrow 0} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} \times \lim _{h \rightarrow 0} \cos \left(x+\frac{h}{2}\right)$
By using the formula we get $-\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
$\therefore f ^{\prime}( x )=2 x \sin x + x ^2 \lim _{h \rightarrow 0} \cos \left(x+\frac{h}{2}\right)$
substuite the value of $h$ to evaluate the limit:
Therefore, $f^{\prime}(x)=2 x \sin x+x^2 \cos (x+0)=2 x \sin x+x^2 \cos x$
Hence,
Derivative of $f(x)=\left(x^2 \sin x\right)$ is $\left(2 x \sin x+x^2 \cos x\right)$