Question

$\frac{d}{d x} (sin x) =$

$cos x$
$- sin x$
$- cos x$
$tan x$

Answer : $cos x$

Solution -

Derivative of $sin x$ is $cos x$

\begin{array}{l} Hence, \frac{d}{d x} (sin x) = cos x \end{array}

Other Way :-

\begin{aligned} Let f (x) & = & sin x, then \\ \frac{d f (x)}{d x} & = & lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = & lim_{h \to 0} \frac{sin (x + h) - sin x}{h} \\ = & lim_{h \to 0} \frac{2 cos (\frac{x + x + h}{2}) sin (\frac{x + h - x}{2})}{h} (∵ sin C - sin D = 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2})) \\ = & lim_{h \to 0} \frac{2 cos (\frac{2 x + h}{2}) sin (\frac{h}{2})}{h} \\ = & lim_{h \to 0} cos (x + \frac{h}{2}) \cdot \frac{sin (\frac{h}{2})}{\frac{h}{2}} \\ = & lim_{h \to 0} cos (x + \frac{h}{2}) \cdot 1 (∵ lim_{x \to 0} \frac{sin x}{x} = 1) \\ = & cos (x + \frac{0}{2}) \\ = & cos (x + 0) \\ = & cos x \end{aligned}

So, correct answer is option-1, $\to cos x$

#Differentiation #BSEB PYQ 2017 #Mathematics #BSEB PYQs