Need to find $dy$ on $dx$ for $x^2+y^3-2y=3$

Question

Need help - as I am not to sure how to do this Above.

I just need an example so that I can do it.

Answer 1

Hints:

1. Try implicit differentiation

2. Apply the chain rule,

i.e, $\frac d {dx}(y^3) = \frac d {dy}(y^3) \cdot \frac{dy}{dx}$

Answer 2

$x^2+y^3−2y=3$ --->(A)

By differentiating (A) with respect to x, we get

$2x + 3y^2 \cdot \frac{dy}{dx} - 2 \frac{dy}{dx} = 0$

$or, \frac{dy}{dx} = \frac{-2x}{3y^2 - 2}$

Slope of the tangent line at the point (2,1) = $\frac {dy}{dx} at (x,y)=(2,1) = \frac{-2*2}{3*1^2 - 2} = \frac{-4}{1} = -4$