Recipe

Find dy/dx by implicit differentiation from F(x,y)=0

Use when: an equation relates x and y but solving explicitly for y is hard or impossible, and you need the slope dy/dx at a specific point

Goal: a numeric slope dy/dx at the given point, using only the two partial derivatives of F

Taught in Implicit differentiation in several variables

1. 1

   Move every term to one side so the equation reads "(some expression in x and y) =0." Call that expression F(x,y).

   Example

   x^2y+y-5=0 is already in this form, with F(x,y)=x^2y+y-5.

   Why: the formula dy/dx=-F_x/F_y only applies once the equation is written this way.

2. 2

   Find F_x (treat y as a constant) and F_y (treat x as a constant).

   Example

   F_x=2xy and F_y=x^2+1.

   Why: these two ordinary partial derivatives are the only ingredients the formula needs, with no separate chain-rule bookkeeping on y required this way.

3. 3

   Substitute the point's coordinates into F_x and F_y.

   Example

   At (2,1): F_x(2,1)=2(2)(1)=4, and F_y(2,1)=2^2+1=5, both defined and F_y\neq0.

   • F_y comes out nonzero — continue to the last step as usual.
   • F_y comes out exactly zero — the formula is undefined there; the curve has a vertical tangent at that point instead of a finite slope.

   Why: F_y=0 is exactly the condition the Implicit Function Theorem excludes, the one place this shortcut formula cannot promise a differentiable y=g(x) nearby.

4. 4

   Divide the negative of F_x's value by F_y's value, both from the previous step.

   Example

   \dfrac{dy}{dx}=-\dfrac{4}{5} at (2,1).

   Why: this ratio is what falls out of differentiating F(x,y(x))=0 with the multivariable chain rule and solving for dy/dx, derived once so it never needs re-deriving per problem.