frac{d}{dx}[e^{ax} cos(bx + c)] = ? | vedclass

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Question

(A) To find the derivative of $e^{ax} \cos(bx + c)$ with respect to $x$,we use the product rule: $\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{

Vedclass · Accepted Answer

$e^{ax}[a \cos(bx + c) - b \sin(bx + c)]$

Vedclass · Answer

(A) To find the derivative of $e^{ax} \cos(bx + c)$ with respect to $x$,we use the product rule: $\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}$.Let $u = e^{ax}$ and $v = \cos(bx + c)$.Then $\frac{du}{dx} = a e^{ax}$ and $\frac{dv}{dx} = -\sin(bx + c) \cdot b = -b \sin(bx + c)$.Applying the product rule:$\frac{d}{dx}[e^{ax} \cos(bx + c)] = e^{ax} \cdot (-b \sin(bx + c)) + \cos(bx + c) \cdot (a e^{ax})$.Factoring out $e^{ax}$,we get:$= e^{ax}[a \cos(bx + c) - b \sin(bx + c)]$.Thus,the correct option is $A$.

$\frac{d}{dx}[e^{ax} \cos(bx + c)] = ?$

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