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Mathematics Question on Differentiability

If f(x)=sin (sin x)f(x) = sin\, (sin\, x) and f′′(x)+tan x f′(x)+g(x)=0,f'' (x) + tan \,x\,f'(x)+ g(x) = 0, then g(x)g(x) is :

A

cos2 x cos (sin x)cos^2\, x\, cos\, (sin\, x)

B

sin2 x cos (cos x)sin^2\, x\, cos\, (cos\, x)

C

sin2 x sin (cos x)sin^2\, x\, sin\, (cos\, x)

D

cos2 x sin (sin x)cos^2\, x\, sin\, (sin\, x)

Answer

cos2 x sin (sin x)cos^2\, x\, sin\, (sin\, x)

Explanation

Solution

f(x)=sin (sin x)f \left(x\right) = sin\, \left(sin\,x\right)
⇒f′(x)=cos(sin x).cos x\Rightarrow f '\left(x\right)=cos\left(sin\,x\right).cos\,x
⇒f′′(x)=−sin (sin x).cos2 cos(sin x).(−sin x)\Rightarrow f ''\left(x\right)=-sin\,\left(sin\,x\right).cos^{2}\,cos\left(sin\,x\right).\left(-sin\,x\right)
=−cos2 x.sin(sin x)−sin x.cos(sin x)=-cos^{2}\,x. sin\left(sin\,x\right)-sin\,x.cos\left(sin\,x\right)
Now f′′(x)+tanx.f′(x)+g(x)=0f ''\left(x\right) + tan x .f '\left(x\right) + g \left(x\right) = 0
⇒g(x)=cos2 x.sin(sin x)+sin x.cos(sin x)−tan x.cos x.cos(sin x)\Rightarrow g\left(x\right) = cos^{2}\, x . sin \left(sin\, x\right) + sin\, x . cos \left(sin\, x\right)- tan \,x . cos\, x . cos \left(sin\, x\right)
⇒g(x)=cos2 x.sin(sin x).\Rightarrow g\left(x\right) = cos^{2}\,x . sin \left(sin\, x\right).