A) g is differentiable at x = 0 and g'(0) = -sin(log2)
B) g is not differentiable at x = 0
C) g'(0) = cos(log2)
D) g'(0) = - cos(log2)
Correct Answer: C
Solution :
In the neighborhoods of x = 0, f(x) = log2 ? sinx \[\therefore \]\[g(x)=f(f(x))=\log 2-\sin (f(x))\] \[=\log 2-\sin (\log 2-\sin x)\] It is differentiable at x = 0, so \[\therefore \]\[g'(x)=-\cos (\log 2-\sin x)(-\cos x)\] \[\therefore \]\[g'(x)=\cos (\log 2)\]
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