If y = frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x | vedclass

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

Question

(C) Given $y = \frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}}$.Using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$,where $u = e^{

Vedclass · Accepted Answer

$\frac{-8}{(e^{2x} - e^{-2x})^2}$

Vedclass · Answer

(C) Given $y = \frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}}$.Using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$,where $u = e^{2x} + e^{-2x}$ and $v = e^{2x} - e^{-2x}$.Then $u' = 2e^{2x} - 2e^{-2x} = 2(e^{2x} - e^{-2x})$ and $v' = 2e^{2x} + 2e^{-2x} = 2(e^{2x} + e^{-2x})$.$\frac{dy}{dx} = \frac{(e^{2x} - e^{-2x}) \cdot 2(e^{2x} - e^{-2x}) - (e^{2x} + e^{-2x}) \cdot 2(e^{2x} + e^{-2x})}{(e^{2x} - e^{-2x})^2}$$= \frac{2(e^{2x} - e^{-2x})^2 - 2(e^{2x} + e^{-2x})^2}{(e^{2x} - e^{-2x})^2}$$= \frac{2[(e^{4x} - 2 + e^{-4x}) - (e^{4x} + 2 + e^{-4x})]}{(e^{2x} - e^{-2x})^2}$$= \frac{2[-4]}{(e^{2x} - e^{-2x})^2} = \frac{-8}{(e^{2x} - e^{-2x})^2}$.

If $y = \frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}}$,then $\frac{dy}{dx} = $

Similar Questions

If $f(x) = \sum_{p=1}^7 p^2 \sin^{-1}\left(\frac{4}{5} \sin(px) - \frac{3}{5} \cos(px)\right)$,then the value of $\frac{df}{dx}$ at $x = 1$ is (Given that $\sin^{-1}(\sin x) = x$)

If $f(2)=4$ and $f^{\prime}(2)=1$,then $\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$ is equal to

Suppose a differentiable function $f(x)$ satisfies the identity $f(x+y) = f(x) + f(y) + xy^2 + x^2y$ for all real $x$ and $y$. If $\lim_{x \rightarrow 0} \frac{f(x)}{x} = 1$,then $f'(3)$ is equal to:

If $f(x) = \frac{1}{1 + \frac{1}{x}}$ and $g(x) = \frac{1}{1 + \frac{1}{f(x)}}$,then $g^{\prime}(2)$ is equal to

If $f(x) = mx + c$,$f(0) = 1$,and $f'(0) = 1$,then find the value of $f(2)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module