Guides / Derivatives

Implicit differentiation: dy/dx on every y term, then solve for dy/dx

Some curves, like a circle, aren't a single function y = f(x), and some equations can't be solved for y at all. Implicit differentiation finds the slope anyway: differentiate the equation as it stands, then solve for dy/dx. Nearly every mistake is a missing chain rule on a y term.

Why it works

Near most points, a curve like x² + y² = 25 is the graph of some function y(x), even if you never write it down. Near (3, 4) it's the top half of the circle, y = √(25 − x²). Plug that function in, and the equation is true for every x nearby, so the derivatives of the two sides are equal. Differentiating y² is then the chain rule on (y(x))²: 2y · dy/dx.

That's also why the answer usually contains y. The same x can sit on more than one branch of the curve (at x = 3, both (3, 4) and (3, −4) are on the circle), and the slope is different on each one. The y in the formula says which branch you're on.

Example 1: a circle

Find the slope of x² + y² = 25 at (3, 4), and the tangent line there.

1. 2x + 2y dydx = 0Differentiate both sides in x. The y² term picks up dy/dx; the constant 25 gives 0.
2. dydx = −xySolve for dy/dx.
3. At (3, 4): −34Plug in both coordinates.
4. y − 4 = −34(x − 3)Point-slope form. At (3, −4) the same formula gives +3/4, the other branch.

Example 2: a product of x and y

Find dy/dx for x²y + y³ = 10 at (1, 2).

1. 1 · 2 + 8 = 10Check the point is on the curve.
2. 2xy + x² dydx + 3y² dydx = 0x²y is a product: the derivative of x² times y, plus x² times the derivative of y. Then y³ gives 3y² dy/dx.
3. (x² + 3y²) dydx = −2xyKeep the dy/dx terms on the left, move the rest, and factor.
4. dydx = −2xyx² + 3y²Divide.
5. At (1, 2): −41 + 12 = −413Plug in x = 1, y = 2.

Example 3: dy/dx on both sides

Find the slope of the folium x³ + y³ = 6xy at (3, 3).

1. 3x² + 3y² dydx = 6y + 6x dydxThe right side is a product, 6x times y.
2. (3y² − 6x) dydx = 6y − 3x²dy/dx terms to the left, the rest to the right.
3. dydx = 2y − x²y² − 2xDivide, and cancel the common factor 3.
4. At (3, 3): 6 − 99 − 6 = −127 + 27 = 54 = 6 · 9, so (3, 3) is on the curve.

Example 4: the second derivative

Find d²y/dx² for x² + y² = 25 at (3, 4).

1. dydx = −xyFrom Example 1.
2. d²ydx² = −y − x · dy/dxy²Quotient rule on x/y. The bottom is y, so its derivative is dy/dx, not 1.
3. = −y + x²/yy² = −x² + y²y³ = −25y³Substitute dy/dx = −x/y, multiply top and bottom by y, and use the original equation.
4. At (3, 4): −2564Negative: the top half of the circle curves downward.

Common mistakes, and the exact line they happen on

1. Leaving dy/dx off a y term

Find dy/dx for x²y + y³ = 10 at (1, 2) (Example 2).

1. 2xy + x² dydx + 3y² = 0This is the mistake. y³ was differentiated as if y were the variable, giving 3y². But y depends on x, so it's 3y² · dy/dx. Solving this line gives dy/dx = −(4 + 12)/1 = −16 instead of −4/13. Every y term gets a dy/dx; every x term doesn't.

2. Skipping the product rule

The same curve, x²y + y³ = 10.

1. 2x dydx + 3y² dydx = 0This is the mistake. x²y was differentiated factor by factor, 2x times dy/dx, as if the derivative of a product were the product of the derivatives. This forces dy/dx = 0 at (1, 2), a flat tangent where the curve is actually falling. The product rule gives 2xy + x² dy/dx.

3. Treating y as a constant in the second derivative

Find d²y/dx² for x² + y² = 25 at (3, 4) (Example 4).

1. dydx = −xyRight so far.
2. d²ydx² = −1y = −14This is the mistake. −x/y was differentiated as if y were a constant. y still depends on x, so the quotient rule brings in dy/dx, and the answer is −25/64.

Practice

1. Find the slope of x² + y² = 169 at (5, 12).

   Show answer

   2x + 2y · dy/dx = 0, so dy/dx = −x/y = −5/12.

2. Find the slope of xy = 6 at (2, 3).

   Show answer

   Product rule: y + x · dy/dx = 0, so dy/dx = −y/x = −3/2.

3. Find the slope of y² = x³ at (4, 8).

   Show answer

   2y · dy/dx = 3x², so dy/dx = 3x²/(2y) = 48/16 = 3.

4. Find the slope of x² − xy + y² = 7 at (1, 3).

   Show answer

   2x − y − x · dy/dx + 2y · dy/dx = 0, so dy/dx = (y − 2x)/(2y − x) = 1/5.

5. If sin y = x with −π/2 < y < π/2, find dy/dx in terms of x.

   Show answer

   cos y · dy/dx = 1, so dy/dx = 1/cos y. On that interval cos y = √(1 − sin²y) = √(1 − x²), so dy/dx = 1/√(1 − x²): the derivative of arcsin x.