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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $A$.

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

Similar Questions

Find the derivative of $2x - \frac{3}{4}$.

If $y = x^2 \log x + \frac{2}{\sqrt{x}},$ then $\frac{dy}{dx} = $

Let $f: (-1, 1) \to R$ be a differentiable function with $f(0) = -1$ and $f'(0) = 1$. Let $g(x) = [f(2f(x) + 2)]^2$. Then $g'(0) = $

If $f(x)=\operatorname{cosec}^{-1}\left[\frac{10}{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}\right]$,then $f^{\prime}(x)$ is equal to:

If $y=\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\log \sqrt{1-x^2}$,then $\frac{d y}{d x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module