Question

Find dy/dx at (1,2) of the function xy²-y-4x²+2=0

Answer 1

$$xy^2-y-4x^2+2=0$$

Differentiate both sides with respect to $x$

$${d\over dx}(xy^2-y-4x^2+2)={d\over dx}(0)$$

Use the Chain rule

$$y^2+x(2y){dy\over dx}-{dy\over dx}-8x+0=0$$

Solve for $\dfrac{dy}{dx}$

$$\dfrac{dy}{dx}=\dfrac{8x-y^2}{2xy-1}$$

Substitute $x=1, y=2$

$$\dfrac{dy}{dx}\big|_{(1,2)}=\dfrac{8(1)-(2)^2}{2(1)(2)-1}=\dfrac{4}{3}$$

$\dfrac{dy}{dx}\big|_{(1,2)}=\dfrac{4}{3}$