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

Differentiate the function with respect to $x$: $(\log x)^{x}+x^{\log x}$

If $y = \sqrt {\frac{{1 + x}}{{1 - x}}} ,$ then $\frac{{dy}}{{dx}} = $

Differentiate the function with respect to $x$: $\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$

If $y=(x+3)^2 \cdot(x+4)^3 \cdot(x+5)^4$,then,the first order derivative of $y$ with respect to $x$ is . . . . . . .

Differentiate the function with respect to $x$: $(x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module