Implicit Differentiation Formula

Implicit differentiation is finding dy/dx when y is defined implicitly by an equation like F(x, y) = 0, by differentiating both sides and solving for.

The Formula

For F(x,y)=0F(x, y) = 0: differentiate both sides with respect to xx, apply chain rule to yy-terms, solve for dydx\frac{dy}{dx}.

When to use: Sometimes you can't (or don't want to) solve for yy explicitly. Instead, differentiate the whole equation as-is. Every time you differentiate a yy-term, attach dydx\frac{dy}{dx} by the chain rule (since yy secretly depends on xx), then solve for dydx\frac{dy}{dx}.

Quick Example

Find dydx\frac{dy}{dx} for the circle x2+y2=25x^2 + y^2 = 25.
Differentiate: 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0.
Solve: dydx=โˆ’xy\frac{dy}{dx} = -\frac{x}{y}
At (3,4)(3, 4): slope =โˆ’34= -\frac{3}{4}.

Notation

dydx\frac{dy}{dx} found implicitly. Alternatively, dydx=โˆ’FxFy\frac{dy}{dx} = -\frac{F_x}{F_y} where FxF_x and FyF_y are partial derivatives of F(x,y)F(x,y).

What This Formula Means

Finding dydx\frac{dy}{dx} when yy is defined implicitly by an equation like F(x,y)=0F(x, y) = 0, by differentiating both sides and solving for dydx\frac{dy}{dx}.

Sometimes you can't (or don't want to) solve for yy explicitly. Instead, differentiate the whole equation as-is. Every time you differentiate a yy-term, attach dydx\frac{dy}{dx} by the chain rule (since yy secretly depends on xx), then solve for dydx\frac{dy}{dx}.

Formal View

If F(x,y)=0F(x, y) = 0 defines yy implicitly as a differentiable function of xx, then by the chain rule: โˆ‚Fโˆ‚x+โˆ‚Fโˆ‚yโ‹…dydx=0\frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \cdot \frac{dy}{dx} = 0, so dydx=โˆ’Fx(x,y)Fy(x,y)\frac{dy}{dx} = -\frac{F_x(x,y)}{F_y(x,y)} provided Fy(x,y)โ‰ 0F_y(x,y) \neq 0.

Worked Examples

Example 1

easy
Find dydx\frac{dy}{dx} for the circle x2+y2=25x^2 + y^2 = 25 and evaluate it at the point (3,4)(3, 4).

Answer

dydx=โˆ’xy\dfrac{dy}{dx} = -\dfrac{x}{y}; at (3,4)(3, 4): slope =โˆ’34= -\dfrac{3}{4}

First step

1
Differentiate both sides with respect to xx: 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0.

Full solution

  1. 2
    Solve for dydx\frac{dy}{dx}: dydx=โˆ’xy\frac{dy}{dx} = -\frac{x}{y}.
  2. 3
    At (3,4)(3, 4): dydx=โˆ’34\frac{dy}{dx} = -\frac{3}{4}.
Whenever a yy-term is differentiated, attach dydx\frac{dy}{dx} by the chain rule. Then collect all dydx\frac{dy}{dx} terms and solve. The tangent to a circle at (3,4)(3,4) has slope โˆ’3/4-3/4.

Example 2

hard
Find dydx\frac{dy}{dx} for x3+y3=6xyx^3 + y^3 = 6xy (folium of Descartes).

Example 3

medium
Find dydx\frac{dy}{dx} for the circle x2+y2=25x^2 + y^2 = 25.

Common Mistakes

  • Differentiating a yy-term without attaching dydx\frac{dy}{dx} - ddx(y3)=3y2dydx\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}, because yy depends on xx.
  • Forgetting the product rule on mixed xyxy-terms - ddx(xy)=y+xdydx\frac{d}{dx}(xy)=y+x\frac{dy}{dx}, not just dydx\frac{dy}{dx}.
  • Leaving dydx\frac{dy}{dx} unsolved - after differentiating, collect all dydx\frac{dy}{dx} terms and solve for it explicitly.

Why This Formula Matters

Many real curves (circles, ellipses, x3+y3=6xyx^3+y^3=6xy) cannot be solved for yy, so implicit differentiation is the only way to get their slopes โ€” and it is the engine behind related rates. It also cements the chain rule: forgetting the dydx\frac{dy}{dx} tag is the telltale sign a student is still thinking of yy as independent. Recognizing it by "Is yy tied to xx by an equation I can't easily solve for yy, and do I need its derivative?" โ€” rather than by familiar numbers โ€” is what lets a student tell it apart from explicit differentiation and chain rule and related rates in a mixed problem set.

Frequently Asked Questions

What is the Implicit Differentiation formula?

Finding dydx\frac{dy}{dx} when yy is defined implicitly by an equation like F(x,y)=0F(x, y) = 0, by differentiating both sides and solving for dydx\frac{dy}{dx}.

How do you use the Implicit Differentiation formula?

Sometimes you can't (or don't want to) solve for yy explicitly. Instead, differentiate the whole equation as-is. Every time you differentiate a yy-term, attach dydx\frac{dy}{dx} by the chain rule (since yy secretly depends on xx), then solve for dydx\frac{dy}{dx}.

What do the symbols mean in the Implicit Differentiation formula?

dydx\frac{dy}{dx} found implicitly. Alternatively, dydx=โˆ’FxFy\frac{dy}{dx} = -\frac{F_x}{F_y} where FxF_x and FyF_y are partial derivatives of F(x,y)F(x,y).

Why is the Implicit Differentiation formula important in Math?

Many real curves (circles, ellipses, x3+y3=6xyx^3+y^3=6xy) cannot be solved for yy, so implicit differentiation is the only way to get their slopes โ€” and it is the engine behind related rates. It also cements the chain rule: forgetting the dydx\frac{dy}{dx} tag is the telltale sign a student is still thinking of yy as independent. Recognizing it by "Is yy tied to xx by an equation I can't easily solve for yy, and do I need its derivative?" โ€” rather than by familiar numbers โ€” is what lets a student tell it apart from explicit differentiation and chain rule and related rates in a mixed problem set.

What do students get wrong about Implicit Differentiation?

The procedure for implicit differentiation is the easy part; the trap is differentiating a yy-term without attaching dydx\frac{dy}{dx}. Asking "Is yy tied to xx by an equation I can't easily solve for yy, and do I need its derivative?" first is what keeps a correct-looking calculation from being attached to the wrong concept.

What should I learn before the Implicit Differentiation formula?

Before studying the Implicit Differentiation formula, you should understand: derivative, chain rule, differentiation rules.

Want the Full Guide?

This formula is covered in depth in our complete guide:

Derivatives Explained: Rules, Interpretation, and Applications โ†’
โ† Back to Implicit Differentiation See All Examples Practice Problems