find derivative of
[tex]y = 2^{sin \: x} \: + \: sin \: (ln \: x)[/tex]
by chain rule.
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find derivative of
[tex]y = 2^{sin \: x} \: + \: sin \: (ln \: x)[/tex]
by chain rule.
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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.
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given function is
[tex]\rm :\longmapsto\:y = {2}^{sinx} + sin(logx)[/tex]
On differentiating both sides w. r. t. x, we get
[tex]\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}\bigg[{2}^{sinx} + sin(logx)\bigg][/tex]
We know,
[tex] \boxed{ \bf{ \: \dfrac{d}{dx}(u + v) = \dfrac{d}{dx}u + \dfrac{d}{dx}v}}[/tex]
So, using this, we get
[tex]\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{d}{dx} {2}^{sinx} + \dfrac{d}{dx}sin(logx)[/tex]
We know,
[tex] \boxed{ \bf{ \: \dfrac{d}{dx} {a}^{x} = {a}^{x}loga}}[/tex]
and
[tex] \boxed{ \bf{ \: \dfrac{d}{dx}sinx = cosx}}[/tex]
So, using this result, we get
[tex]\rm :\longmapsto\:\dfrac{dy}{dx} = {2}^{sinx} \: \dfrac{d}{dx}sinx +cos(logx)\dfrac{d}{dx}logx[/tex]
We know,
[tex] \boxed{ \bf{ \: \dfrac{d}{dx}logx = \frac{1}{x}}}[/tex]
So, using this, we get
[tex]\rm :\longmapsto\:\dfrac{dy}{dx} = {2}^{sinx} \: cosx +cos(logx) \times \dfrac{1}{x}[/tex]
[tex]\bf :\longmapsto\:\dfrac{dy}{dx} = cosx \: {2}^{sinx} \: + \: \dfrac{cos(logx)}{x} [/tex]
Additional Information :-
[tex]\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}[/tex]