Question

If $y=e^{ax}\cdot \cos (bx+c)$, then find $\frac{dy}{dx}$

• A. $a{e}^{ax}\cos (bx+c)-b{e}^{ax}\sin (bx+c)$
• B. $a{e}^{ax}\cos (bx+c)+b{e}^{ax}\sin (bx+c)$
• C. $ea{e}^{ax}\cos (bx+c)-b{e}^{ax}\sin (bx+c)$
• D. $a{e}^{ax}\cos (bx+c)$

Answer 1

The correct option is A $a{e}^{ax}\cos (bx+c)-b{e}^{ax}\sin (bx+c)$

$y=e^{ax}\cdot \cos (bx+c)$

On differentiating, we get

$\frac{dy}{dx}=e^{ax}\frac{d}{dx}\left(\cos (bx+c)\right)+\cos (bx+c)\frac{d}{dx}{e}^{ax}$

$=-e^{ax}\sin (bx+c)\frac{d}{dx}(bx+c)+\cos (bx+c)e^{ax}\frac{d}{dx}\left(ax\right)$

$=-b{e}^{ax}\sin (bx+c)+a\cos (bx+c)e^{ax}$

$=a{e}^{ax}\cos (bx+c)-b{e}^{ax}\sin (bx+c)$