Q. If $y = x^{2}e^{mx},\ where\ m\ is\ a\ constant,\ then\frac{d^{3}y}{dx^{3}}$ is equal to

A) $me^{mx}(m^{2}x^{2} + 6mx + 6)$
B) $2m^{3}xe^{mx}$
C) $me^{mx}(m^{2}x^{2} + 2mx + 2)$
D) None of these
Explanation

(a)
Given, $y = x^{2}e^{mx} \Rightarrow \frac{dy}{dx} = 2xe^{mx} + mx^{2}e^{mx}$
$\Rightarrow \ \frac{d^{2}y}{dx^{2}} = 2\left( e^{mx} + mxe^{mx} \right) + m(2xe^{mx} - x^{2}me^{mx})\$
$\Rightarrow \frac{d^{2}y}{dx^{2}} = e^{mx}\left( m^{2}x^{2} + 4mx + 2 \right)$
$\Rightarrow \frac{d^{3}y}{dx^{3}} = e^{mx}\lbrack m^{3}x^{2} + 4m^{2}x + 2m + 2m^{2}x + 4m\rbrack$
$= e^{mx}\lbrack m^{3}x^{2} + 6m^{2}x + 6m\rbrack$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.