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 $A$.

List-$I$	List-$II$
$A$. $\sec^{-1} x$	$I$. $\frac{1}{1-x^2}, x \in (-1, 1)$
$B$. $\tanh^{-1} x$	$II$. $\frac{-1}{\|x\| \sqrt{x^2+1}}, x \neq 0$
$C$. $\coth^{-1} x$	$III$. $\frac{1}{\|x\| \sqrt{x^2-1}}, \|x\| > 1$
$D$. $\operatorname{cosech}^{-1} x$	$IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$
	$V$. $\frac{-1}{\|x\| \sqrt{1-x^2}}, \|x\| < 1, x \neq 0$

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

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