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