Correct Answer - Option 3 :
\(\frac{ e^{sinx}(1+x \cdot cosx \cdot logx)}{x}\)
Concept:
Derivative of logx with respect to x is 1/x
Derivative of ex with respect to x is ex
Product rule: (uv)' = uv' + vu'
Calculation:
Let y = esinx ⋅ log x
We differentiate the function with respect to x
As we know that, (uv)' = uv' + vu'
⇒ y' = esinx [logx]' + [logx] (esinx)'
⇒ \(\rm y' = e^{sinx}\times \frac{1}{x}+log(x).e^{sinx}.(sinx)'\)
⇒ \(\rm y' = \frac{ e^{sinx}}{x}+log(x).e^{sinx}.(cosx)\)
⇒ \(\rm y' = \frac{ e^{sinx}}{x}+e^{sinx}.cosx.log(x)\)
⇒ \(\rm y' = \frac{ e^{sinx}(1+x \cdot cosx \cdot logx)}{x}\)
Hence, option 3 is correct.