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Find the derivative of esinx ⋅ log x with respect to x ?
1. \(\rm \frac{ e^{sinx}(1+x \cdot cosx \cdot logx)}{x^2}\)
2. \(\rm \frac{ e^{sinx}(1+cosx \cdot logx)}{x}\)
3. \(\frac{ e^{sinx}(1+x \cdot cosx \cdot logx)}{x}\)
4. \(\rm \frac{ e^{sinx}(1-x \cdot cosx \cdot logx)}{x}\)

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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.

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