Suppose that f(x) is a differentiable function such that f'(x) is continuous f'(0)=1 and f''(0) does not exist. If g(x)=xf'(x). Then,
g'(0) does not exist
g'(0)=0
g'(0)=1
g'(0)=2
Explanation for the correct option:
Given:
f'(0)=1
g(x)=xf'(x)
By Differentiating it, we get
g'(x)=f'(x)+xf''(x) ∵ddx(uv)=udvdx+vdudx
⇒g'(0)=f'(0)+0f''(0)
∴g'(0)=1
Hence, Option ‘C’ is Correct.