If y = frac{e^x log x}{x^2},then frac{dy}{dx} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given,$y = \frac{e^x \log x}{x^2}$.Applying the quotient rule for differentiation,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx}

Vedclass · Accepted Answer

$\frac{e^x \{1 + (x - 2) \log x\}}{x^3}$

Vedclass · Answer

(D) Given,$y = \frac{e^x \log x}{x^2}$.Applying the quotient rule for differentiation,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.Here,$u = e^x \log x$ and $v = x^2$.$\frac{du}{dx} = e^x \log x + e^x \cdot \frac{1}{x} = e^x \left( \log x + \frac{1}{x} \right)$.$\frac{dv}{dx} = 2x$.Therefore,$\frac{dy}{dx} = \frac{x^2 \left[ e^x \left( \log x + \frac{1}{x} \right) \right] - (e^x \log x)(2x)}{(x^2)^2}$.$\frac{dy}{dx} = \frac{x^2 e^x \log x + x e^x - 2x e^x \log x}{x^4}$.$\frac{dy}{dx} = \frac{x e^x \log x + e^x - 2 e^x \log x}{x^3}$.$\frac{dy}{dx} = \frac{e^x [x \log x + 1 - 2 \log x]}{x^3}$.$\frac{dy}{dx} = \frac{e^x [1 + (x - 2) \log x]}{x^3}$.Thus,option $D$ is correct.

If $y = \frac{e^x \log x}{x^2}$,then $\frac{dy}{dx} =$

Similar Questions

Differentiate the function with respect to $x$: $\sin^{-1}(x\sqrt{x})$,where $0 \le x \le 1$.

If $f(x)=3 e^{x^2}$ then $f^{\prime}(x)-2 x f(x)+\frac{1}{3} f(0)-f^{\prime}(0)=$

If $f(1) = 1$ and $f'(1) = 3$,then the derivative of $f(f(f(x))) + (f(x))^2$ at $x = 1$ is:

Let $f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3)$,where $x \in R$. Then $f'(10)$ is equal to:

The values of $x$,at which the first derivative of the function $(\sqrt{x} + \frac{1}{\sqrt{x}})^2$ with respect to $x$ is $\frac{3}{4}$,are

Vedclass Test Series

Exam Paper Generator

Online Exam Module