$cos (x^{2} + 1)$

$(Differentiate with respect to x, using first principle)$

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Solution

Find derivative by first principle method

Let $f (x) = cos (x^{2} + 1)$

$f' (x) = {lim}_{h \to 0} \frac{cos [(x + h)^{2} + 1] - cos (x^{2} + 1)}{h}$

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

$= {lim}_{h \to 0} \frac{2 sin ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{\begin{matrix} 2 x^{2} + h^{2} + 2 x h + 2 \end{matrix}}{2} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ sin (\frac{- h^{2} - 2 x h}{2})}{h}$
$[∵ cos C - cos D = 2 sin \frac{C + D}{2} sin \frac{D - C}{2}]$

$= {lim}_{h \to 0} \frac{- 2 sin ⎛ ⎜ ⎝ \begin{matrix} x^{2} + x h + \frac{h^{2}}{2} + 1 \end{matrix} ⎞ ⎟ ⎠ sin (\frac{2 x h + h^{2}}{2})}{h}$

$= {lim}_{h \to 0} - 2 sin ⎛ ⎜ ⎝ \begin{matrix} x^{2} + x h + \frac{h^{2}}{2} + 1 \end{matrix} ⎞ ⎟ ⎠ . {lim}_{h \to 0} \frac{sin \frac{(h (2 x + h)}{2}}{h}$

$= - 2 sin (x^{2} + 1) {lim}_{h \to 0} \frac{sin [\frac{h (2 x + h)}{2}]}{h . h (\frac{2 x + h}{2})}$

$\times \frac{h (2 x + h)}{2}$

$= - 2 sin (x^{2} + 1) {lim}_{h \to 0} \frac{sin [\frac{h (2 x + h)}{2}]}{h (\frac{2 x + h}{2})}$ $\times {lim}_{h \to 0} \frac{(2 x + h)}{2}$
$= - 2 sin (x^{2} + 1) . \frac{2 x}{2}$ $(∵ {lim}_{θ \to 0} \frac{sin θ}{θ} = 1)$

$= - 2 x sin (x^{2} + 1)$

Hence the required answer is

$- 2 x sin (x^{2} + 1)$

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