The derivative of (log x)^{x} with respect to | vedclass

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

Question

(B) Let $u = (\log x)^x$. Taking logarithm on both sides,we get $\log u = x \log(\log x)$.Now,differentiating with respect to $x$,we have $\frac{1}{u}

Vedclass · Accepted Answer

$x(\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right]$

Vedclass · Answer

(B) Let $u = (\log x)^x$. Taking logarithm on both sides,we get $\log u = x \log(\log x)$.Now,differentiating with respect to $x$,we have $\frac{1}{u} \frac{du}{dx} = x \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log(\log x) \cdot 1 = \frac{1}{\log x} + \log(\log x)$.Thus,$\frac{du}{dx} = (\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right]$.Let $v = \log x$. Then $\frac{dv}{dx} = \frac{1}{x}$.We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{(\log x)^x [\frac{1}{\log x} + \log(\log x)]}{1/x} = x(\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right]$.

The derivative of $(\log x)^{x}$ with respect to $\log x$ is

Similar Questions

If $y = ((x+1)(4x+1)(9x+1) \ldots (n^2x+1))^2$,then $\frac{dy}{dx}$ at $x=0$ is

If $y = [(x+1)(2x+1)(3x+1) \ldots (nx+1)]^4$,then $\frac{dy}{dx}$ at $x=0$ is

Differentiate the function with respect to $x$: $\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

If $y = x^{\sin x}$,then $\frac{dy}{dx} = $

If $y = ({x^x})^x$,then $\frac{dy}{dx} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module