Question

if f(1)=-2 ,f'(1)=1, Find the derivative of logf(e^x) wrt x at x=0

Answer 1

assuming log = ln(e^x)

Answer 2

\[\frac{d}{dx}\log(f(e^x))=\frac{f'(e^x)\times e^x}{f(e^x)}\] by the chain rule.

Answer 3

so at

\[x=0\] you get
\[\frac{d}{dx}\log(f(e^0))=\frac{f'(e^0)\times e^0}{f(e^0)}=\frac{f'(1)\times 1}{f(1)}\] so now plug in the values you know

Answer 4

oh ok Thankyou :)