Correct Answer - Option 2 :
\(\rm xcotx+log(sinx)\)
Concept:
Derivative of logx with respect to x is 1/x
Product rule: (uv)' = uv' + vu'
Calculation:
Let y = x log(sinx)
We differentiate the function with respect to x
As we know that, (uv)' = uv' + vu'
⇒ y' = x [log(sinx)]' + [log(sinx)] (x)'
⇒ \(\rm y' = x\times \frac{1}{sinx}(sinx)'+log(sinx)1\)
⇒ \(\rm y' = x\times \frac{1}{sinx}(cosx)+log(sinx)\)
⇒ \(\rm y' = x\times \frac{cosx}{sinx}+log(sinx)\)
⇒ \(\rm y' = xcotx+log(sinx)\)
Hence, option 2 is correct.
