If $y=xy^2+4x^2, $ then $\frac{dy}{dx}$ is equal to __________.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{8x+y^2}{1-2xy}$
$y=xy^2+4x^2$ $[(uv),=u'v+v'u]$
$⇒\frac{dy}{dx}=y^2+2xy\frac{dy}{dx}+8x$
$⇒\frac{dy}{dx}(1-2xy)=8x+y^2$
$⇒\frac{dy}{dx}=\frac{8x+y^2}{1-2xy}$