The derivative of e^{3x} sin 4x with respect | vedclass

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(A) Let $y = e^{3x} \sin 4x$. Applying the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$: $\frac{dy}{dx} = e^{3x} \frac{d}{dx}(\

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$5 e^{3x} \sin \left(4x + 	an^{-1} \frac{4}{3}ight)$

Vedclass · Answer

(A) Let $y = e^{3x} \sin 4x$. Applying the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$: $\frac{dy}{dx} = e^{3x} \frac{d}{dx}(\sin 4x) + \sin 4x \frac{d}{dx}(e^{3x})$ $\frac{dy}{dx} = e^{3x} (4 \cos 4x) + \sin 4x (3 e^{3x})$ $\frac{dy}{dx} = e^{3x} (3 \sin 4x + 4 \cos 4x)$ To simplify $3 \sin 4x + 4 \cos 4x$,we multiply and divide by $\sqrt{3^2 + 4^2} = 5$: $\frac{dy}{dx} = 5 e^{3x} \left( \frac{3}{5} \sin 4x + \frac{4}{5} \cos 4x ight)$ Let $\cos \alpha = \frac{3}{5}$ and $\sin \alpha = \frac{4}{5}$,so $	an \alpha = \frac{4}{3}$,which means $\alpha = 	an^{-1} \frac{4}{3}$. Then $\frac{dy}{dx} = 5 e^{3x} (\cos \alpha \sin 4x + \sin \alpha \cos 4x)$ Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$: $\frac{dy}{dx} = 5 e^{3x} \sin(4x + \alpha) = 5 e^{3x} \sin \left(4x + 	an^{-1} \frac{4}{3}ight)$.

The derivative of $e^{3x} \sin 4x$ with respect to $x$ is:

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