frac{d^n}{dx^n}(e^{2x} + e^{-2x}) = | vedclass

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

Question

(C) Let $y = e^{2x} + e^{-2x}$.First derivative: $\frac{dy}{dx} = 2e^{2x} - 2e^{-2x} = 2(e^{2x} - e^{-2x})$.Second derivative: $\frac{d^2y}{dx^2} = 4e

Vedclass · Accepted Answer

$2^n[e^{2x} + (-1)^n e^{-2x}]$

Vedclass · Answer

(C) Let $y = e^{2x} + e^{-2x}$.First derivative: $\frac{dy}{dx} = 2e^{2x} - 2e^{-2x} = 2(e^{2x} - e^{-2x})$.Second derivative: $\frac{d^2y}{dx^2} = 4e^{2x} + 4e^{-2x} = 2^2(e^{2x} + e^{-2x})$.Third derivative: $\frac{d^3y}{dx^3} = 8e^{2x} - 8e^{-2x} = 2^3(e^{2x} - e^{-2x})$.Observing the pattern,the $n$-th derivative is given by $\frac{d^n}{dx^n}(e^{2x} + e^{-2x}) = 2^n[e^{2x} + (-1)^n e^{-2x}]$.Thus,the correct option is $C$.

$\frac{d^n}{dx^n}(e^{2x} + e^{-2x}) = $

Similar Questions

If $y = \cos^{2} \frac{3x}{2} - \sin^{2} \frac{3x}{2}$,then $\frac{d^{2}y}{dx^{2}}$ is

For $n \in N$,if $y = a x^{n+1} + b x^{-n}$,then $x^2 \frac{d^2 y}{d x^2} = $

If $y=44 x^{45}+45 x^{-44}$, then $y^{\prime \prime}=$

If $x = \log p$ and $y = \frac{1}{p}$,then

$\frac{d^2}{dx^2}(2\cos x \cos 3x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module