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

Differentiate the function with respect to $x$: $\sin^{3} x + \cos^{6} x$

Differentiate the function with respect to $x$: $\cos(x^{3}) \cdot \sin^{2}(x^{5})$

$y=\log \left\{\left(\frac{1+x}{1-x}\right)^{1 / 4}\right\}-\frac{1}{2} \tan ^{-1}(x)$,then $\frac{d y}{d x}$ is equal to

The derivative of $\sqrt{\sqrt{x} + 1}$ is

Let $f: R \to R$ be a function such that $f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3)$,for all $x \in R$. Then $f(2)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module