If y = log_e(log_e x) for x 1,then frac{d^2 | vedclass

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

Question

(C) Given $y = \log_e(\log_e x)$.First,find the first derivative $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{1}{\log_e x} \cdot \frac{

Vedclass · Accepted Answer

$-\frac{\log_e e x}{x^2 (\log_e x)^2}$

Vedclass · Answer

(C) Given $y = \log_e(\log_e x)$.First,find the first derivative $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{1}{\log_e x} \cdot \frac{d}{dx}(\log_e x) = \frac{1}{x \log_e x} = (x \log_e x)^{-1}$.Now,find the second derivative $\frac{d^2 y}{dx^2}$ using the power rule and product rule:$\frac{d^2 y}{dx^2} = -1(x \log_e x)^{-2} \cdot \frac{d}{dx}(x \log_e x)$.$\frac{d^2 y}{dx^2} = -\frac{1}{(x \log_e x)^2} \cdot [1 \cdot \log_e x + x \cdot \frac{1}{x}] = -\frac{\log_e x + 1}{x^2 (\log_e x)^2}$.Since $\log_e x + 1 = \log_e x + \log_e e = \log_e(ex)$,we have:$\frac{d^2 y}{dx^2} = -\frac{\log_e(ex)}{x^2 (\log_e x)^2}$.

If $y = \log_e(\log_e x)$ for $x > 1$,then $\frac{d^2 y}{dx^2} = $ . . . . . . .

Similar Questions

For the function $y = \tan^{-1} x$,the value of $(1 + x^2) y_2$ is . . . . . .

If $m \sin ^{-1} x = \log _{e} y$,then $(1 - x^{2}) y'' - x y'$ is equal to

If $x = A \cos 4t + B \sin 4t$,then $\frac{d^2x}{dt^2} = $

The equation $x^5 - 5x^3 + 5x^2 - 1 = 0$ has how many equal roots?

If $y=2 \sin x+3 \cos x$ and $y+A \frac{d^2 y}{d x^2}=B$,then the values of $A$ and $B$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module