Find the derivative of x sinx from first principle. - Vidyadip

Question

Find the derivative of x sinx from first principle.

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Answer

We have, f(x) = x sinx
By using first principle of derivative,
$f^{\prime}(x)=\underset{{h \rightarrow 0}}{\lim} \frac{f(x+h)-f(x)}{h}$
$\begin{array}{l}=\underset{{h \rightarrow 0}}{\lim} \frac{(x+h) \sin (x+h)-x \sin x}{h} \\ =\underset{{h \rightarrow 0}}{\lim} \frac{(x+h)[\sin x \cdot \cos h+\cos x \cdot \sin h]-x \sin x}{h}[\because \sin ( x + y )=\sin x \cos y + cos x \sin y] \\ =\underset{{h \rightarrow 0}}{\lim} \frac{[x \sin x \cdot \cos h+x \cdot \cos x \cdot \sin h+h \sin x \cdot \cos h+h \cos x \cdot \sin h-x \sin x)}{h} \\ =\underset{{h \rightarrow 0}}{\lim} \frac{\mid x \sin x(\cos h-1)+x \cdot \cos x \cdot \sin h+h(\sin x \cdot \cos h+\cos x \cdot \sin h)]}{h} \\ =\underset{{h \rightarrow 0}}{\lim} \frac{x \sin x(\cos h-1)}{h}+\underset{{h \rightarrow 0}}{\lim} x \cdot \cos x \cdot \frac{\sin h}{h}+\underset{{h \rightarrow 0}}{\lim} \frac{h(\sin x \cdot \cos h+\cos x \cdot \sin h)}{h}\end{array}$
$\begin{array}{l}=x \sin x \underset{{h \rightarrow 0}}{\lim}\left[\frac{-(1-\cos h)}{h}\right]+ x \cos x +\sin x \\ =-2 x \sin x \cdot \lim _{\frac{h}{2} \rightarrow 0} \frac{\sin ^2 \frac{h}{2}}{h \times \frac{h}{4}} \times \frac{h}{4}+ x \cos x +\sin x \\ =-x \cdot \sin x \cdot \frac{2}{4} \lim _{\frac{h}{2} \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^2 \times h + x \cos x +\sin x \\ =-x \sin x \cdot \frac{1}{2}(1) \times 0+ x \cos x +\sin x \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ = x \cos x +\sin x \end{array}$

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