The differential coefficient of f(logx) when f(x)=logx is:
xlogx
1xlogx
logxx
Differentiating to find the coefficient:
Given that: f(x)=logx
Now,
f(logx)=loglogx⇒f'(logx)=1logx×1x∵ddxlogx=1x⇒f'(logx)=1xlogx
Hence, the correct answer is Option (C).
The differential coefficient of ax+logxsinx is: