Question

$x\cos x(\text{Differentiate with respect to}x,\text{using first principle})$

Answer 1

Find derivative by first principle method

Let $f\left(x\right)=x\cos x$

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

$\left[f^{\prime }\right(x)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right))}{h}]$

$={lim}_{h\to 0}\frac{x\cos (x+h)+h\cos (x+h)-x\cos x}{h}$

$={lim}_{h\to 0}\frac{x\left[\begin{array}{c}\cos (x+h)-\\ \cos x\end{array}\right]+h\cos (x+h)}{h}$

$={lim}_{h\to 0}\frac{\begin{array}{c}2x\sin \left(\frac{2x+h}{2}\right)\sin (-\frac{h}{2})+\\ h\cos (x+h)\end{array}}{h}$

$\because \cos C-\cos D=2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{D-C}{2}\right)$

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

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

$=-x\sin x+\cos x$ $(\because {lim}_{\theta \to 0}\frac{\sin \theta }{\theta }=1)$
$\text{hence the required answer is}$

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