Answer:
[tex]\frac{dy}{dx}=\frac{y(1-6x^2)}{(1+y)}[/tex]
Step-by-step explanation:
Given function : 4x³ + ln(y²) + 2y = 2x
on differentiating both sides with respect to 'x', we get
[tex]\frac{d(4x^3 + ln(y^2) + 2y)}{dx}= \frac{d(2x)}{dx}[/tex]
or
[tex]12x^2+\frac{2}{y}(\frac{dy}{dx})+2(\frac{dy}{dx})=2[/tex]
or
[tex]12x^2+(\frac{2}{y}+2)\frac{dy}{dx}=2[/tex]
or
[tex](\frac{2+2y}{y})\frac{dy}{dx}=2-12x^2[/tex]
or
[tex]\frac{dy}{dx}=\frac{y(2-12x^2)}{2+2y}[/tex]
or
[tex]\frac{dy}{dx}=\frac{2y(1-6x^2)}{2(1+y)}[/tex]
or
[tex]\frac{dy}{dx}=\frac{y(1-6x^2)}{(1+y)}[/tex]
