Find the second order derivative of the funct | vedclass

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(A) Let $y = e^{6x} \cos 3x$. Using the product rule,$\frac{dy}{dx} = \frac{d}{dx}(e^{6x}) \cdot \cos 3x + e^{6x} \cdot \frac{d}{dx}(\cos 3x)$. $\frac

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$9 e^{6x}(3 \cos 3x - 4 \sin 3x)$

Vedclass · Answer

(A) Let $y = e^{6x} \cos 3x$. Using the product rule,$\frac{dy}{dx} = \frac{d}{dx}(e^{6x}) \cdot \cos 3x + e^{6x} \cdot \frac{d}{dx}(\cos 3x)$. $\frac{dy}{dx} = 6e^{6x} \cos 3x + e^{6x} (-3 \sin 3x) = 6e^{6x} \cos 3x - 3e^{6x} \sin 3x$. Now,differentiate again with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(6e^{6x} \cos 3x) - \frac{d}{dx}(3e^{6x} \sin 3x)$. $\frac{d^2y}{dx^2} = 6[6e^{6x} \cos 3x - 3e^{6x} \sin 3x] - 3[6e^{6x} \sin 3x + 3e^{6x} \cos 3x]$. $\frac{d^2y}{dx^2} = 36e^{6x} \cos 3x - 18e^{6x} \sin 3x - 18e^{6x} \sin 3x - 9e^{6x} \cos 3x$. $\frac{d^2y}{dx^2} = 27e^{6x} \cos 3x - 36e^{6x} \sin 3x$. Factoring out $9e^{6x}$,we get: $\frac{d^2y}{dx^2} = 9e^{6x}(3 \cos 3x - 4 \sin 3x)$.

Find the second order derivative of the function $y = e^{6x} \cos 3x$.

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