Question

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

Answer 1

Correct Answer - Option 3 : \( \rm \frac{e^{sinx}(xcosx-1)}{x^2}\)

Concept:

Derivative of e^x with respect to x is ex

Quotient rule: \(\rm \left ( \frac{u}{v} \right )'= \frac{vu'-uv'}{v^2}\)

Calculation:

Given function is \( \rm \frac{e^{sinx}}{x}\)

We differentiate the function with respect to x

⇒ \( \rm \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{x(e^{sinx})'-e^{sinx}(x)'}{(x)^2}\)

⇒ \( \rm \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{xe^{sinx}(sinx)'-e^{sinx}(1)}{(x)^2}\)

⇒ \( \rm \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{xe^{sinx}(cosx)-e^{sinx}}{x^2}\)                  

⇒ \( \rm \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{e^{sinx}(xcosx-1)}{x^2}\)

Hence, option 3 is correct.