Q. If $y = 2^{x}.3^{2x - 1}$, then $\frac{dy}{dx}$ is equal to
A) $(\log{2)(\log{3)}}$
B) $(\log{18)}$
C) $(\log{18^{2})y^{2}}$
D) $(\log{18)y}$
Explanation
(d)
Given, $y = 2^{x}.3^{2x - 1}$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx} = 2^{x}.3^{2x - 1}\log{3(2) + 2^{x}.3^{2x - 1}\log 2}$
$\Rightarrow \ \frac{dy}{dx} = 2^{x}.3^{2x - 1}\lbrack 2\log{3 + \log{2\rbrack}}$
$\Rightarrow \ \frac{dy}{dx} = y\log 18$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.