If y = log_2(log_2(x)),then frac{dy}{dx} is e | vedclass

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

Question

(A) Given $y = \log_2(\log_2(x))$.Using the change of base formula $\log_a b = \frac{\log_e b}{\log_e a}$,we can write:$y = \frac{\log_e(\log_2 x)}{\l

Vedclass · Accepted Answer

$\frac{\log_2 e}{x \log_e x}$

Vedclass · Answer

(A) Given $y = \log_2(\log_2(x))$.Using the change of base formula $\log_a b = \frac{\log_e b}{\log_e a}$,we can write:$y = \frac{\log_e(\log_2 x)}{\log_e 2} = \frac{\log_e(\frac{\log_e x}{\log_e 2})}{\log_e 2}$$y = \frac{\log_e(\log_e x) - \log_e(\log_e 2)}{\log_e 2}$Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{d}{dx} [\log_e(\log_e x) - \log_e(\log_e 2)]$$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{1}{\log_e x} \cdot \frac{d}{dx}(\log_e x)$$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{1}{\log_e x} \cdot \frac{1}{x}$Since $\frac{1}{\log_e 2} = \log_2 e$,we get:$\frac{dy}{dx} = \frac{\log_2 e}{x \log_e x}$.

If $y = \log_2(\log_2(x))$,then $\frac{dy}{dx}$ is equal to

Similar Questions

If the two curves $y=a^x$ and $y=b^x$ intersect at an angle $\alpha$,then $\tan \alpha=$

$\frac{d}{d x}\left(\log \left(\frac{1}{x}\right)+\log \left(\frac{1}{x^2}\right)+\log\left(\frac{1}{x^3}\right)\right) = \text{ . . . . . . }$,$x > 1$

If $f(x)=\log _e\left(e^{2 x}\left(\frac{3 x+5}{5-3 x}\right)^{\frac{2}{3}}\right)$,$x \neq \frac{-5}{3}, \frac{5}{3}$,then the value of $\frac{d f}{d x}$ at $x=1$ is

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

The differential coefficient of the function $\log_e \left( \sqrt{\frac{1 + \sin x}{1 - \sin x}} \right)$ with respect to $x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module