Question
Differentiate sin (2x + 3) w.r.t. x from first principle.
$y = \triangle y = \sin (2x +2\triangle x + 3)$
$\therefore \triangle y \sin (2x + 2\triangle x + 3) -\sin (2x + 3)$
$= 2\cos (2x + 3 + \triangle x) \sin \triangle x$
$\therefore \lim\limits_{\triangle x\rightarrow 0} \frac{\triangle y}{\triangle x} = \lim\limits_{\triangle x \rightarrow 0} 2\cos (2x +3 +\triangle x) \lim\limits_{\triangle x\rightarrow 0} \frac {\sin\triangle x}{\triangle x}$
OR
$\frac{dy}{dx} = \text2\cos (2x + 3) 1 = 2\cos (2x + 3)$
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$\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{8-\sin2\text{x}}}\text{ dx}$