Find the derivative of f(x) = x sin x using t | vedclass

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(A) By the first principle of derivatives,$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$.Given $f(x) = x \sin x$,we have $f(x+h) = (x+h) \sin(x+h)$.

Vedclass · Accepted Answer

$x \cos x + \sin x$

Vedclass · Answer

(A) By the first principle of derivatives,$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$.Given $f(x) = x \sin x$,we have $f(x+h) = (x+h) \sin(x+h)$.$f'(x) = \lim_{h 	o 0} \frac{(x+h) \sin(x+h) - x \sin x}{h}$.Using the identity $\sin(x+h) = \sin x \cos h + \cos x \sin h$:$f'(x) = \lim_{h 	o 0} \frac{(x+h)(\sin x \cos h + \cos x \sin h) - x \sin x}{h}$.$f'(x) = \lim_{h 	o 0} \frac{x \sin x \cos h + x \cos x \sin h + h \sin x \cos h + h \cos x \sin h - x \sin x}{h}$.$f'(x) = \lim_{h 	o 0} \left[ x \sin x \left( \frac{\cos h - 1}{h} ight) + x \cos x \left( \frac{\sin h}{h} ight) + \sin x \cos h + \cos x \sin h ight]$.As $h 	o 0$,$\frac{\cos h - 1}{h} 	o 0$,$\frac{\sin h}{h} 	o 1$,$\cos h 	o 1$,and $\sin h 	o 0$.$f'(x) = x \sin x (0) + x \cos x (1) + \sin x (1) + \cos x (0) = x \cos x + \sin x$.

Find the derivative of $f(x) = x \sin x$ using the first principle.

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