Implicit Differentiation

Some curves cannot be solved for y and still have a slope at all but a few of their points. Choose a curve and an x: the page finds every point above that x, draws the tangent at the one you pick, and works out −F_x/F_y there. Then step through the differentiation and check each line against the numbers.

Implicit differentiation	$\frac{d y}{d x} = - \frac{F _{x}}{F _{y}}, F_{y} \neq = 0$

Worked example — Implicit Differentiation

The chosen point	(1, 2.3615)
∂F/∂x at the point	−11.1688
∂F/∂y at the point	10.7296
The slope there	1.0409
Other points above this x	0

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Implicit Differentiation

A curve F(x, y) = 0 that cannot be solved for y still has a slope at each of its smooth points: dy/dx = −F_x/F_y, a formula in x and y together. It breaks where F_y = 0.

What each symbol means

F_x: ∂F/∂x, the derivative of F in x with y held fixed
F_y: ∂F/∂y, the derivative of F in y with x held fixed; where it is 0 the slope formula has no value

Where the formula comes from

Differentiate along the curve. Treat y as a function of x and differentiate F(x, y(x)) = 0. Each x term differentiates as usual; each y term picks up a factor y′ by the chain rule. The two pieces are F_x and F_y·y′, and they add to 0.
$\frac{d}{d x} F (x, y (x)) = F_{x} + F_{y} y^{'} = 0$
Solve for y′. Wherever F_y is not zero, divide by it. The slope comes out in terms of x and y, so it needs the point, not just x: one x can carry several points and several slopes.
$y^{'} = - \frac{F _{x}}{F _{y}} (F_{y} \neq = 0)$
Where F_y = 0. The formula has nothing to divide by. If F_x is not zero there, the tangent is vertical; if F_x is zero too, the formula reads 0/0 and says nothing, as at the folium's crossing.

Assumes: F has partial derivatives at the point. On the three built-in curves the points above an x are found in closed form; on a typed curve they are found numerically, by scanning F along y for a change of sign, so a point where F touches 0 without crossing it is not listed.

Problems solved in full

The tangent to a circle 3 steps
Find dy/dx on x² + y² = 25, and the tangent line at (3, 4).

Open this problem in the tool
1. Differentiate both sides with y a function of x. The chain rule puts a y′ beside the derivative of y²:
  
  $2 x + 2 y y^{'} = 0$
2. Collect y′ and solve:
  
  $y^{'} = - \frac{x}{y}$
3. At (3, 4), then the line through that point with that slope:
  
  $(3, 4) : y^{'} = - \frac{3}{4}; y - 4 = - \frac{3}{4} (x - 3) \Rightarrow y = - \frac{3}{4} x + \frac{25}{4}$
Answer

y′ = −x/y, and the tangent at (3, 4) is y = −3x/4 + 25/4. The page prints F_x = 6 and F_y = 8 there, and −6/8 is the slope. The tangent is at right angles to the radius, whose slope is 4/3.

$y^{'} = - \frac{x}{y}, y = - \frac{3}{4} x + \frac{25}{4}$
A product inside the curve 3 steps
Find dy/dx on x²y + y² = 2 and its value at (1, 1).

Open this problem in the tool
Show solution
1. Differentiate both sides. x²y is a product, so it gives two terms, and every y picks up a y′:
  
  $2 x y + x^{2} y^{'} + 2 y y^{'} = 0$
2. Collect the y′ terms and solve:
  
  $(x^{2} + 2 y) y^{'} = - 2 x y \Rightarrow y^{'} = - \frac{2 x y}{x ^{2} + 2 y}$
3. At (1, 1):
  
  $(1, 1) : y^{'} = - \frac{2}{3}$
Answer

y′ = −2xy/(x² + 2y), and −2/3 at (1, 1). On the page: Your own, x^2*y + y^2 = 2, x = 1. It prints F_x = 2 and F_y = 3.

$y^{'} = - \frac{2 x y}{x ^{2} + 2 y}$
A cube of y 3 steps
Find dy/dx on y³ + xy = 2 and its value at (1, 1).

Open this problem in the tool
Show solution
1. Differentiate both sides. y³ gives 3y²y′ by the chain rule, and xy gives y + xy′ by the product rule:
  
  $3 y^{2} y^{'} + y + x y^{'} = 0$
2. Collect and solve:
  
  $(3 y^{2} + x) y^{'} = - y \Rightarrow y^{'} = - \frac{y}{3 y ^{2} + x}$
3. At (1, 1):
  
  $(1, 1) : y^{'} = - \frac{1}{4}$
Answer

y′ = −y/(3y² + x), and −1/4 at (1, 1). No one can solve this curve for y in a line, and the slope never needed it.

$y^{'} = - \frac{y}{3 y ^{2} + x}$
Where the tilted ellipse is level 3 steps
Find the points of x² + xy + y² = 3 where the tangent is horizontal.

Open this problem in the tool
Show solution
1. Differentiate and solve for y′ as before:
  
  $2 x + y + x y^{'} + 2 y y^{'} = 0 \Rightarrow y^{'} = - \frac{2 x + y}{x + 2 y}$
2. The slope is zero where the numerator is, so put y = −2x back into the curve:
  
  $y^{'} = 0 ⟺ y = - 2 x : x^{2} - 2 x^{2} + 4 x^{2} = 3 \Rightarrow x = \pm 1$
3. The two points:
  
  $(1, - 2) (- 1, 2)$
Answer

(1, −2) and (−1, 2). At (1, −2) the page prints F_x = 0 and a slope of 0. Where the denominator is zero instead, at (2, −1) and (−2, 1), the tangent is vertical.

$(1, - 2), (- 1, 2)$
The derivative of ln x 3 steps
Use eʸ = x to find the derivative of y = ln x.

Open this problem in the tool
Show solution
1. Differentiate both sides of eʸ = x. The left side is a chain, so it gives eʸ times y′:
  
  $e^{y} y^{'} = 1$
2. Solve, and use eʸ = x once more:
  
  $y^{'} = \frac{1}{e ^{y}} = \frac{1}{x}$
3. So y = ln x has slope 1/x:
  
  $y = ln x \Rightarrow \frac{d}{d x} ln x = \frac{1}{x}; x = 2 : \frac{1}{2}$
Answer

(ln x)′ = 1/x. At x = 2 the page prints a slope of 0.5. The argument used only the derivative of eˣ, which is how the rule for ln x is first proved.

$\frac{d}{d x} ln x = \frac{1}{x}$
The second derivative of a circle 3 steps
Find y″ on x² + y² = 25 and its value at (3, 4).

Open this problem in the tool
Show solution
1. Start from the first derivative:
  
  $y^{'} = - \frac{x}{y}$
2. Differentiate again with the quotient rule, replace y′ by −x/y, and use x² + y² = 25:
  
  $y^{''} = - \frac{y - x y ^{'}}{y ^{2}} = - \frac{y + \frac{x ^{2}}{y}}{y ^{2}} = - \frac{x ^{2} + y ^{2}}{y ^{3}} = - \frac{25}{y ^{3}}$
3. At (3, 4):
  
  $(3, 4) : y^{''} = - \frac{25}{64} \approx - 0.3906$
Answer

y″ = −25/y³, and −25/64 ≈ −0.3906 at (3, 4). The second derivative is past what the page computes; it prints the slope −0.75 that the calculation starts from.

$y^{''} = - \frac{25}{y ^{3}}$
The tip of the folium 3 steps
Find the slope of x³ + y³ = 6xy at the tip of its loop, (3, 3).

Open this problem in the tool
Show solution
1. Differentiate both sides. The right side is a product:
  
  $3 x^{2} + 3 y^{2} y^{'} = 6 y + 6 x y^{'}$
2. Collect y′ and solve, dividing through by 3:
  
  $(3 y^{2} - 6 x) y^{'} = 6 y - 3 x^{2} \Rightarrow y^{'} = \frac{2 y - x ^{2}}{y ^{2} - 2 x}$
3. At (3, 3):
  
  $(3, 3) : y^{'} = \frac{6 - 9}{9 - 6} = - 1$
Answer

y′ = −1 at (3, 3). The curve is symmetric in y = x, so its tangent there has to cross that line at right angles. The page prints F_x = F_y = 9.

$y^{'} (3, 3) = - 1$
The top of the leaf 4 steps
Find the highest point of the folium's leaf exactly, and decide whether it is the tip (3, 3), where the leaf meets the line y = x. Press Top of the leaf: at x = 2.52 the upper point's slope reads −0.0002, close to 0, but a slider of step 0.01 can only bracket the answer.

Open this problem in the tool
Show solution
1. The tangent is horizontal where y′ = 0, which needs the numerator 2y − x² to be 0 while the denominator is not:
  
  $y^{'} = \frac{2 y - x ^{2}}{y ^{2} - 2 x} = 0 \Rightarrow y = \frac{x ^{2}}{2}$
2. Put y = x²/2 into the curve. x = 0 is the origin, where the formula is 0/0, so divide by x³:
  
  $x^{3} + \frac{x ^{6}}{8} = 6 x \cdot \frac{x ^{2}}{2} = 3 x^{3} \Rightarrow \frac{x ^{6}}{8} = 2 x^{3} \Rightarrow x^{3} = 16$
3. So x is the cube root of 16, and the denominator there is not 0:
  
  $x = 2^{4/3} = 2.5198, y = \frac{x ^{2}}{2} = 2^{5/3} = 3.1748, y^{2} - 2 x = 2^{10/3} - 2^{7/3} = 2^{7/3} = 5.0397 \neq = 0$
4. At the tip (3, 3) the slope is −1, so the leaf is still rising to the left of it:
  
  $y^{'} (3, 3) = \frac{2 \cdot 3 - 3 ^{2}}{3 ^{2} - 2 \cdot 3} = \frac{- 3}{3} = - 1, 3.1748 - 3 = 0.1748$
Answer

The top of the leaf is (2^4/3, 2^5/3) ≈ (2.5198, 3.1748), higher than the tip by 0.1748. Its height is the x of the vertical tangent in Where it folds, because swapping x and y leaves x³ + y³ = 6xy unchanged and turns horizontal tangents into vertical ones. The page reads a slope wherever you put the point; solving y′ = 0 is algebra it does not do.

$(2^{4/3}, 2^{5/3}) \approx (2.5198, 3.1748)$

Three points above x = 1, and three slopes 🖖

Press Three slopes at x = 1. Above x = 1 the folium has three points, and the curve passes through each with a different slope. A formula in x alone gives one number for each x, so it cannot be this curve's slope. The implicit derivative, y′ = (2y − x²)/(y² − 2x), asks for the point's y as well, and the table under the curve evaluates it at all three. Descartes, who first wrote about the curve in 1638, believed the leaf repeated in every quadrant like the petals of a flower. There is one leaf, and two tails.

The formula fails exactly where the curve gains or loses a point 🖖

Press Where it folds. Here the denominator F_y is zero and the tangent is vertical. To the left there are three points above x; here the upper two meet; to the right one is left. On all three curves the number of points above x changes exactly where F_y = 0, and nowhere else. That is the implicit function theorem seen from the slider: where F_y ≠ 0, the curve near a point is the graph of one function, and where F_y = 0, two of those functions run into each other.

At the origin the formula says 0/0, and the curve has two tangents 🖖

Press The crossing. At the origin both F_x and F_y are zero, so the formula reads 0/0, which is more than a vertical tangent. The curve passes through this point twice: once leaving flat along the leaf's lower edge, once rising steeply across it. Approach the origin along x = 0.1, 0.01, 0.001 and the table shows one slope shrinking towards 0 and two growing without bound. No single slope could be right, and the canvas draws both tangents.

Learning path

Differentiation: how rates combine

References (2)

The method as the page steps it (differentiate both sides with y a function of x, then solve for dy/dx), the circle x² + y² = 25 with its tangent at (3, −4) (Example 3.71), and the folium in the form x³ + y³ − 3xy = 0 with slope −1 at (3/2, 3/2) (Example 3.72), the a = 1 version of the page's tip (3, 3): G. Strang and E. Herman, Calculus Volume 1, OpenStax (2016), section 3.8, “Implicit Differentiation”.
That Descartes first discussed the folium in 1638 and believed the leaf was repeated in each quadrant like the petals of a flower (insight block 1), and the asymptote x + y + a = 0 (here a = 2): J. J. O'Connor and E. F. Robertson, “Folium of Descartes”, Famous Curves Index, MacTutor History of Mathematics Archive, University of St Andrews.

By the 42Math team · Reviewed against published sources. Corrections welcome.