If y=e^{4x} cos 5x,then frac{d^{2}y}{dx^{2}} | vedclass

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(A) Given $y = e^{4x} \cos 5x$. Applying the product rule,$\frac{dy}{dx} = e^{4x}(-5 \sin 5x) + \cos 5x(4e^{4x}) = e^{4x}(4 \cos 5x - 5 \sin 5x)$. Now

Vedclass · Accepted Answer

$-9$

Vedclass · Answer

(A) Given $y = e^{4x} \cos 5x$. Applying the product rule,$\frac{dy}{dx} = e^{4x}(-5 \sin 5x) + \cos 5x(4e^{4x}) = e^{4x}(4 \cos 5x - 5 \sin 5x)$. Now,differentiating again with respect to $x$: $\frac{d^{2}y}{dx^{2}} = e^{4x}(-20 \sin 5x - 25 \cos 5x) + (4 \cos 5x - 5 \sin 5x)(4e^{4x})$. Evaluating at $x=0$: $\left(\frac{d^{2}y}{dx^{2}}\right)_{x=0} = e^{0}(-20 \sin 0 - 25 \cos 0) + (4 \cos 0 - 5 \sin 0)(4e^{0})$. Since $\sin 0 = 0$ and $\cos 0 = 1$: $\left(\frac{d^{2}y}{dx^{2}}\right)_{x=0} = 1(0 - 25) + (4 - 0)(4) = -25 + 16 = -9$.

If $y=e^{4x} \cos 5x$,then $\frac{d^{2}y}{dx^{2}}$ at $x=0$ is

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