The derivative of e^{ax} cos bx with respect | vedclass

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

Question

(A) Let $y = e^{ax} \cos bx$.Using the product rule,the derivative is:$\frac{dy}{dx} = \frac{d}{dx}(e^{ax}) \cdot \cos bx + e^{ax} \cdot \frac{d}{dx}(

Vedclass · Accepted Answer

$\sqrt{a^{2}+b^{2}}$

Vedclass · Answer

(A) Let $y = e^{ax} \cos bx$.Using the product rule,the derivative is:$\frac{dy}{dx} = \frac{d}{dx}(e^{ax}) \cdot \cos bx + e^{ax} \cdot \frac{d}{dx}(\cos bx)$$\frac{dy}{dx} = ae^{ax} \cos bx - be^{ax} \sin bx$$\frac{dy}{dx} = e^{ax} (a \cos bx - b \sin bx)$Let $a = r \cos \alpha$ and $b = r \sin \alpha$.Then $a^2 + b^2 = r^2(\cos^2 \alpha + \sin^2 \alpha) = r^2$.Thus,$r = \sqrt{a^2 + b^2}$.Substituting these into the derivative expression:$\frac{dy}{dx} = e^{ax} (r \cos \alpha \cos bx - r \sin \alpha \sin bx)$$\frac{dy}{dx} = re^{ax} (\cos bx \cos \alpha - \sin bx \sin \alpha)$Using the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$,we get:$\frac{dy}{dx} = re^{ax} \cos(bx + \alpha)$Therefore,$r = \sqrt{a^2 + b^2}$.

The derivative of $e^{ax} \cos bx$ with respect to $x$ is $re^{ax} \cos(bx + \alpha)$,where $\alpha = \tan^{-1}(\frac{b}{a})$. When $a > 0, b > 0$,the value of $r$ is:

Similar Questions

If $a$ and $b$ are fixed non-zero constants,then the derivative of $\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x$ is $ma+nb-p$,where

The value of $\Delta \log f(x) + \Delta^{2}(3^{x})$ is

Find the derivative of $f(x) = 3$ at $x = 0$ and at $x = 3$.

If $f(x)=\cosh ^{-1}\left(\frac{1-x}{1+x}\right)$ is well defined, then $f^{\prime}(x)=$

Find the derivative of the following function (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\frac{\sin (x+a)}{\cos x}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module