Question

find derivative of
$y = 2^{sin \: x} \: + \: sin \: (ln \: x)$
by chain rule.​

Answer 1

Answer:

ans 2

Step-by-step explanation:

find derivative of

y = 2^{sin \: x} \: + \: sin \: (ln \: x)y=2

sinx

+sin(lnx)

by chain rule.

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:y = {2}^{sinx} + sin(logx)$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}\bigg[{2}^{sinx} + sin(logx)\bigg]$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx}(u + v) = \dfrac{d}{dx}u + \dfrac{d}{dx}v}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{d}{dx} {2}^{sinx} + \dfrac{d}{dx}sin(logx)$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx} {a}^{x} = {a}^{x}loga}}$

and

$\boxed{ \bf{ \: \dfrac{d}{dx}sinx = cosx}}$

So, using this result, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {2}^{sinx} \: \dfrac{d}{dx}sinx +cos(logx)\dfrac{d}{dx}logx$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx}logx = \frac{1}{x}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {2}^{sinx} \: cosx +cos(logx) \times \dfrac{1}{x}$

$\bf :\longmapsto\:\dfrac{dy}{dx} = cosx \: {2}^{sinx} \: + \: \dfrac{cos(logx)}{x}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$