frac{d}{dx} log(log x) = | vedclass

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

Question

(C) To find the derivative of $\log(\log x)$ with respect to $x$,we use the chain rule.Let $u = \log x$. Then the expression becomes $\frac{d}{dx} \lo

Vedclass · Accepted Answer

$(x \log x)^{-1}$

Vedclass · Answer

(C) To find the derivative of $\log(\log x)$ with respect to $x$,we use the chain rule.Let $u = \log x$. Then the expression becomes $\frac{d}{dx} \log(u)$.By the chain rule,$\frac{d}{dx} \log(u) = \frac{d}{du} \log(u) \cdot \frac{du}{dx}$.We know that $\frac{d}{du} \log(u) = \frac{1}{u}$ and $\frac{du}{dx} = \frac{d}{dx} \log x = \frac{1}{x}$.Substituting these values back,we get $\frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$.This can be written as $(x \log x)^{-1}$.Therefore,the correct option is $C$.

$\frac{d}{dx} \log(\log x) = $

Similar Questions

$\frac{d}{dx}\left( \frac{1}{x^4 \sec x} \right) = $

The value of the integral $\int_{-1}^{+1} \frac{1}{t^3} \, dt$ is:

If $y = x \sin x$,then

The value of $1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\ldots$ up to $\infty$ is

If $\tan \theta = \frac{1}{\sqrt{5}}$ and $\theta$ lies in the first quadrant,the value of $\cos \theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module