Differentiate (x^ 2 + 1)with respect to xfrom first principle. - Vidyadip

Question

Differentiate $\sin(x^{2} + 1)$with respect to $ x$from first principle.

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Answer

$\frac{\triangle y}{\triangle x} = \frac {\sin \bigg[( {x + \triangle x)^{2}}+ 1\bigg] -\sin ( x^{2} + 1)}{\triangle x} $$\therefore \frac{dy}{dx} = \lim\limits_{\triangle x \rightarrow 0} \frac{2\cos\bigg[ \frac{( x + \triangle x)^{2} + x^{2}}{2}\bigg] .\sin \bigg[\frac{\triangle x( 2x + \triangle x}{2} \bigg]} {\triangle x}$
$= 2.\cos ( x^{2} + 1) . \lim\limits_{\triangle x \rightarrow 0} \frac{\sin\bigg[\frac{(\triangle x+{2 x} +\triangle{x)}}{2}\bigg] \bigg[\frac{(2x +\triangle x)}{2}\bigg]}{\triangle x.\bigg(\frac{2x + \triangle x}{2}\bigg)}$
$= \text{2 x} .\cos ( x^{2} + 1)$

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