A) \[{{\cos }^{2}}x\cos (\sin x)\]
B) \[{{\sin }^{2}}x\cos (\cos x)\]
C) \[{{\sin }^{2}}x\sin (\cos x)\]
D) \[{{\cos }^{2}}x\sin (\sin x)\]
Correct Answer: D
Solution :
\[f(x)=\sin (\sin x)\] \[\Rightarrow \]\[f'(x)=\cos (\sin x).\cos x\] \[\Rightarrow \]\[f''(x)=-\sin (\sin x).{{\cos }^{2}}x+\cos (\sin x).(-\sin x)\] \[\Rightarrow \]\[=-{{\cos }^{2}}x.\sin (\sin x)-\sin x.\cos (\sin x)\] Now\[f''(x)+\tan x.f'(x)+g(x)=0\] \[\Rightarrow \]\[g(x)={{\cos }^{2}}x.\sin (\sin x)+\sin x.\cos (\sin x)\] Now\[f''(x)+\tan x.f'(x)+g(x)=0\] \[\Rightarrow \]\[g(x)={{\cos }^{2}}x.\sin (\sin x)+\sin x.\cos (\sin x)\] \[-\tan x.\cos x.\cos (\sin x)\] \[\Rightarrow \]\[g(x)={{\cos }^{2}}x.\sin (\sin x).\]You need to login to perform this action.
You will be redirected in
3 sec