Find the derivative: frac{d}{dx} left[ frac{e | vedclass

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(C) To find the derivative of $\frac{e^{ax}}{\sin(bx + c)}$,we use the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx}

Vedclass · Accepted Answer

$\frac{e^{ax}[a\sin(bx + c) - b\cos(bx + c)]}{\sin^2(bx + c)}$

Vedclass · Answer

(C) To find the derivative of $\frac{e^{ax}}{\sin(bx + c)}$,we use the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.Let $u = e^{ax}$ and $v = \sin(bx + c)$.Then $\frac{du}{dx} = a e^{ax}$ and $\frac{dv}{dx} = b \cos(bx + c)$.Applying the quotient rule:$\frac{d}{dx} \left[ \frac{e^{ax}}{\sin(bx + c)} \right] = \frac{\sin(bx + c) \cdot (a e^{ax}) - e^{ax} \cdot (b \cos(bx + c))}{\sin^2(bx + c)}$.Factoring out $e^{ax}$ in the numerator:$= \frac{e^{ax}[a \sin(bx + c) - b \cos(bx + c)]}{\sin^2(bx + c)}$.

Find the derivative: $\frac{d}{dx} \left[ \frac{e^{ax}}{\sin(bx + c)} \right]$

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