Q. If $y = x - x^{2},\ then\ the\ derivatives\ of\ y^{2}\ w.r.t.x^{2}$ is
A) $2x^{2} + 3x - 1$
B) $2x^{2} - 3x + 1$
C) $2x^{2} + 3x + 1$
D) $2x^{2} - 3x - 1$
Explanation
(b)
$\because\ y = x - x^{2}$
$\therefore\ \ \frac{dy}{dx} = 1 - 2x$
Now,
$\frac{d\left( y^{2} \right)}{d(x^{2})} = \frac{\frac{d\left( y^{2} \right)}{dx}}{\frac{d\left( x^{2} \right)}{dx}}$
$= \frac{2y\frac{dy}{dx\ }}{2x}$
$= \frac{y}{x}(1 - 2x)$
$= (1 - x)(1 - 2x)\ \$
= $1 - 3x + 2x^{2}$
= $2x^{2} - 3x + 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.