Q. If $y = a^{x}.b^{2x - 1\ },\ then\frac{d^{2}y}{dx^{2}}\$is
A) $y^{2}\log{ab^{2}}$
B) $y\log{ab^{2}}$
C) $y^{2}$
D) $y{(\log{ab^{2})}}^{2}$
Explanation
(d)
Given, $y = a^{x}.b^{2x - 1}$
$\Rightarrow \log{y = x\log{a + (2x - 1)\log b}}\ \ \ \ \ \ \ \ \ \ \ \$
$\Rightarrow \frac{1}{y}.\frac{dy}{dx} = \log{a + 2\log b}$
$\Rightarrow \frac{dy}{dx} = y\log{ab^{2}}$
$\Rightarrow \frac{d^{2}y}{dx^{2}} = \frac{dy}{dx}\log{ab^{2} = y\ {(\log{ab^{2})}}^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.