Differentiate the following with respect to x | vedclass

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

Question

(A) Let $y = \log (\log x)$.By using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} [\log (\log x)]$Applying the ch

Vedclass · Accepted Answer

$\frac{1}{x \log x}$

Vedclass · Answer

(A) Let $y = \log (\log x)$.By using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} [\log (\log x)]$Applying the chain rule,where the derivative of $\log(u)$ is $\frac{1}{u} \cdot \frac{du}{dx}$:$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{d}{dx} (\log x)$Since the derivative of $\log x$ is $\frac{1}{x}$:$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{1}{x}$Thus,the final result is:$\frac{dy}{dx} = \frac{1}{x \log x}$,for $x > 1$.

Differentiate the following with respect to $x$: $\log (\log x)$,where $x > 1$.

Similar Questions

If $y = \log_{\sin x} (\tan x)$,then $\left( \frac{dy}{dx} \right)_{\pi/4}$ is equal to

Mrs. Rajni deposited $Rs. 10,000$ in a bank that pays $4 \%$ interest compounded continuously. The amount she gets after $10$ years is approximately $Rs.$ . . . . . . . (Given $e^{(0.4)} = 1.49182$)

If the angle between the curves $y = a^x$ and $y = b^x$ is $\alpha$,find the modulus of $\tan \alpha$.

If $\phi(x) = \log_{5} \log_{3} x$,then $\phi'(e)$ is equal to

If $y = f \left( \frac{3 + 2x}{3 - 2x} \right)$, where $f(x) = \tan(\log x)$ and $\frac{dy}{dx} = \left( \frac{A}{B + Cx^2} \right) \cdot \sec^2 \left( \log \left( \frac{3 + 2x}{3 - 2x} \right) \right)$, then the respective values of $A, B$ and $C$ are ......

Vedclass Test Series

Exam Paper Generator

Online Exam Module