Question

The derivative of $e^{ax}\cos bx$ with respect to $x$ is $r{e}^{ax}$ $\cos (bx-Ta{n}^{-1}\frac{b}{a})$ When $a>0,b>0$,the value of $r$ is ............

• A. $ab$
• B. $a+b$
• C. $\sqrt{a^2+b^2}$
• D. $\frac{1}{\sqrt{ab}}$

Answer 1

Answer: (C)

Given, $\frac{d}{dx}\left(e^{ax}\cos bx\right)=r{e}^{ax}\cos (bx+\alpha )$, then
$r=?$
Let $y=e^{ax}\cdot \cos bx$
Let $y=e^{ax}\cdot \cos bx$
$\frac{dy}{dx}=a{e}^{ax}\cdot \cos bx-b{e}^{ax}\cdot \sin bx$
$\frac{dy}{dx}=e^{ax}(a\cos bx-b\sin bx)$
$a=r\cos \alpha <br/><br/>b=r\sin \alpha \}$...(i)
Then, $\frac{dy}{dx}=e^{ax}\cdot r\{\cos bx\cdot \cos \alpha -\sin bx\cdot \sin \alpha \}$
$\frac{dy}{dx}=e^{ax}\cdot r\cos (bx+\alpha )$...(ii)
Where, $\tan \alpha =\frac{b}{a}\Rightarrow \alpha ={\tan }^{-1}\left(\frac{b}{a}\right)$
and $r^2\left({\cos }^2\alpha +{\sin }^2\alpha \right)=a^2+b^2$
$r^2=a^2+b^2$
$r=\sqrt{a^2+b^2}$