Answer to Question #244998 in Calculus for Susan

Question

Use the implicit differentiation formula to find dy/dx for the equation arctan(x^2 y) = x + xy^2

Answer 1

Given that $tan^{-1} (x^2y)=x+xy^2$

Differentiating both sides, we get:

$\frac{1}{1+x^4y^2}\times (2xy+x^2\frac{dy}{dx})=1+y^2+2xy\frac{dy}{dx}\\$

$\Rightarrow \frac{2xy}{1+x^4y^2}+\frac{x^2}{1+x^4y^2}\frac{dy}{dx}=1+y^2+2xy\frac{dy}{dx}\\$

$\Rightarrow \frac{dy}{dx}(\frac{x^2}{1+x^4y^2}-2xy)=1+y^2-\frac{2xy}{1+x^4y^2}\\ \Rightarrow \frac{dy}{dx}(\frac{x^2-2xy-2x^5y^2}{1+x^4y^2})=\frac{1+x^4y^2+y^2+x^4y^4-2xy}{1+x^4y^2}\\ \Rightarrow \frac{dy}{dx}=\frac{1+x^4y^2+y^2+x^4y^4-2xy}{x^2-2xy-2x^5y^2}$