If f(x) = frac{e^{2x} - e^{-2x}}{e^{3x} + e^{ | vedclass

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

Question

(D) Given $f(x) = \frac{e^{2x} - e^{-2x}}{e^{3x} + e^{-3x}}$.Using the quotient rule $\left( \frac{u}{v} \right)^{\prime} = \frac{u^{\prime}v - uv^{\p

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Given $f(x) = \frac{e^{2x} - e^{-2x}}{e^{3x} + e^{-3x}}$.Using the quotient rule $\left( \frac{u}{v} \right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$:Let $u = e^{2x} - e^{-2x}$ and $v = e^{3x} + e^{-3x}$.Then $u^{\prime} = 2e^{2x} + 2e^{-2x}$ and $v^{\prime} = 3e^{3x} - 3e^{-3x}$.At $x = 0$:$u(0) = e^0 - e^0 = 1 - 1 = 0$.$v(0) = e^0 + e^0 = 1 + 1 = 2$.$u^{\prime}(0) = 2(1) + 2(1) = 4$.$v^{\prime}(0) = 3(1) - 3(1) = 0$.Now, $f^{\prime}(0) = \frac{u^{\prime}(0)v(0) - u(0)v^{\prime}(0)}{(v(0))^2}$.$f^{\prime}(0) = \frac{(4)(2) - (0)(0)}{(2)^2} = \frac{8}{4} = 2$.

If $f(x) = \frac{e^{2x} - e^{-2x}}{e^{3x} + e^{-3x}}$, then $f^{\prime}(0) = $

Similar Questions

$\frac{d}{d x}\left(e^{\log _e \sqrt{1+\tan ^2 x}}\right) =$

If $f(x)=\operatorname{cosec}^{-1}\left[\frac{10}{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}\right]$,then $f^{\prime}(x)$ is equal to:

$\frac{d}{d x}\left(\sqrt{\frac{1-\tan x}{1+\tan x}}\right) = $

Let $y = \left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x + \log_{e}(1+x)$ for $x > -1$. Then, at $x = 0$, $\frac{dy}{dx}$ equals:

If $y = f\left( \frac{5x + 1}{10x^2 - 3} \right)$ and $f'(x) = \cos x$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module