The derivative of e^{ax} cos bx with respect | vedclass

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(A) Let $y = e^{ax} \cos bx$.Using the product rule,the derivative is:$\frac{dy}{dx} = \frac{d}{dx}(e^{ax}) \cdot \cos bx + e^{ax} \cdot \frac{d}{dx}(

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$\sqrt{a^{2}+b^{2}}$

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(A) Let $y = e^{ax} \cos bx$.Using the product rule,the derivative is:$\frac{dy}{dx} = \frac{d}{dx}(e^{ax}) \cdot \cos bx + e^{ax} \cdot \frac{d}{dx}(\cos bx)$$\frac{dy}{dx} = ae^{ax} \cos bx - be^{ax} \sin bx$$\frac{dy}{dx} = e^{ax} (a \cos bx - b \sin bx)$Let $a = r \cos \alpha$ and $b = r \sin \alpha$.Then $a^2 + b^2 = r^2(\cos^2 \alpha + \sin^2 \alpha) = r^2$.Thus,$r = \sqrt{a^2 + b^2}$.Substituting these into the derivative expression:$\frac{dy}{dx} = e^{ax} (r \cos \alpha \cos bx - r \sin \alpha \sin bx)$$\frac{dy}{dx} = re^{ax} (\cos bx \cos \alpha - \sin bx \sin \alpha)$Using the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$,we get:$\frac{dy}{dx} = re^{ax} \cos(bx + \alpha)$Therefore,$r = \sqrt{a^2 + b^2}$.

The derivative of $e^{ax} \cos bx$ with respect to $x$ is $re^{ax} \cos(bx + \alpha)$,where $\alpha = \tan^{-1}(\frac{b}{a})$. When $a > 0, b > 0$,the value of $r$ is:

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