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

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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}) = $

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