Implicit Differentiation

A curve that is not y = f(x)

$x^2+y^2=25$x² + y² = 25 is the circle of radius 5 centered at the origin. Solving for y gives two answers, $y=\sqrt{25-x^2}$y = √(25 − x²) for the top half and $y=-\sqrt{25-x^2}$y = −√(25 − x²) for the bottom half. Between x = −5 and x = 5 each x has two points above it, so no single function of x gives the whole circle.

Implicit differentiation finds $\frac{dy}{dx}$dy/dx from the equation as it stands, without choosing a half. The equation defines y implicitly, as a function of x near each point, even where no formula for it is written down.

Differentiating a term in y

Differentiate every term of $x^2+y^2=25$x² + y² = 25 with respect to x, treating y as a function of x. $x^2$x² gives 2x, and the constant 25 gives 0.

$y^2$y² is a function inside a function: the inner function is y and the outer function is the square. So the chain rule gives $2y\times \frac{dy}{dx}$2y × dy/dx. The power rule on y gives 2y, and then the inner derivative, $\frac{dy}{dx}$dy/dx, multiplies it. Every term in y gains a factor $\frac{dy}{dx}$dy/dx in the same way: $y^3$y³ gives $3y^2\frac{dy}{dx}$3y² dy/dx.

Solving for $\frac{dy}{dx}$dy/dx

So $2x+2y\frac{dy}{dx}=0$2x + 2y dy/dx = 0. Take 2x from both sides: $2y\frac{dy}{dx}=-2x$2y dy/dx = −2x. Divide both sides by 2y: $\frac{dy}{dx}=-\frac{x}{y}$dy/dx = −x/y, wherever $y\ne 0$y ≠ 0.

At (3, 4) the gradient is $-\frac{3}{4}$−3/4, and at (3, −4) it is $\frac{3}{4}$3/4. At (5, 0), y = 0 and $-\frac{x}{y}$−x/y has no value: the tangent there is vertical.

Check on the top half, $y=\sqrt{25-x^2}$y = √(25 − x²). At x = 3.001, y = 3.9992498, so the chord from (3, 4) has gradient $-\frac{0.0007502}{0.001}=-0.7502$−0.0007502/0.001 = −0.7502, close to $-\frac{3}{4}$−3/4. The chain rule on $\sqrt{25-x^2}$√(25 − x²) gives $-\frac{x}{\sqrt{25-x^2}}$−x/√(25 − x²), which is $-\frac{x}{y}$−x/y again.

The geometry agrees. The radius to (3, 4) has gradient $\frac{4}{3}$4/3, and $\left(\frac{4}{3}\right)\times \left(-\frac{3}{4}\right)=-1$(4/3) × (−3/4) = −1, so the tangent is perpendicular to the radius, as a tangent to a circle must be.

The circle $x^2+y^2=25$x² + y² = 25. The gold line is the tangent at (3, 4), with gradient $-\frac{x}{y}=-\frac{3}{4}$−x/y = −3/4; the dashed line through the origin carries the radius, with gradient $\frac{4}{3}$4/3. They meet at a right angle.

A point held on the circle $x^2+y^2=25$x² + y² = 25, with the tangent there drawn at gradient $-\frac{x}{y}$−x/y. Where x and y have the same sign the tangent falls, where they differ it rises, at the top and bottom it is flat, and at the sides it is vertical.

A term with both x and y

In xy = 12, the term xy is a product of x and y, so it needs the product rule: $1\times y+x\times \frac{dy}{dx}$1 × y + x × dy/dx. So $y+x\frac{dy}{dx}=0$y + x dy/dx = 0, and $\frac{dy}{dx}=-\frac{y}{x}$dy/dx = −y/x. At (3, 4), which is on the curve since 3 × 4 = 12, the gradient is $-\frac{4}{3}$−4/3.

Here the equation can also be solved: $y=\frac{12}{x}$y = 12/x, so $\frac{dy}{dx}=-\frac{12}{x^2}$dy/dx = −12/x², which is $-\frac{12}{9}=-\frac{4}{3}$−12/9 = −4/3 at x = 3.

A curve that cannot be rearranged

$y^5+y=x$y⁵ + y = x has exactly one y for each x, but no formula gives it. Differentiate anyway: $5y^4\frac{dy}{dx}+\frac{dy}{dx}=1$5y⁴ dy/dx + dy/dx = 1. Factor out $\frac{dy}{dx}$dy/dx: $\left(5y^4+1\right)\frac{dy}{dx}=1$(5y⁴ + 1) dy/dx = 1, so $\frac{dy}{dx}=\frac{1}{5y^4+1}$dy/dx = 1/(5y⁴ + 1). This is never 0 and always positive, so the curve rises everywhere.

At (2, 1), which is on the curve since 1 + 1 = 2, the gradient is $\frac{1}{6}=0.1667$1/6 = 0.1667. Check: solving $y^5+y=2.001$y⁵ + y = 2.001 numerically gives y = 1.00016662, so the chord from (2, 1) has gradient 0.16662. At the origin the gradient is $\frac{1}{0+1}=1$1/(0 + 1) = 1.

The gold curve $y^5+y=x$y⁵ + y = x, drawn by solving for y at each x. At (2, 1) its tangent, the gold line, has gradient $\frac{1}{5y^4+1}=\frac{1}{6}$1/(5y⁴ + 1) = 1/6; at the origin the curve leaves at gradient 1.

The method

Differentiate every term with respect to x. A term in y alone gains a factor $\frac{dy}{dx}$dy/dx, by the chain rule. A term with both x and y needs the product rule. A constant gives 0. Then collect the terms in $\frac{dy}{dx}$dy/dx on one side, factor out $\frac{dy}{dx}$dy/dx, and divide.

For $x^2+xy+y^2=7$x² + xy + y² = 7, the terms give 2x, then $y+x\frac{dy}{dx}$y + x dy/dx, then $2y\frac{dy}{dx}$2y dy/dx, and 0. So $\left(x+2y\right)\frac{dy}{dx}=-\left(2x+y\right)$(x + 2y) dy/dx = −(2x + y), and $\frac{dy}{dx}=-\frac{2x+y}{x+2y}$dy/dx = −(2x + y)/(x + 2y). At (1, 2), which is on the curve since 1 + 2 + 4 = 7, the gradient is $-\frac{4}{5}=-0.8$−4/5 = −0.8. Solving the equation numerically for y near x = 1 gives the same slope.

The usual mistakes

Differentiating $y^2$y² as 2y. That differentiates with respect to y; with respect to x, the chain rule adds the factor $\frac{dy}{dx}$dy/dx.

Differentiating a product such as 9xy as $9\frac{dy}{dx}$9 dy/dx. The product rule gives $9y+9x\frac{dy}{dx}$9y + 9x dy/dx.

Dropping a coefficient: for $3x^2+y^2=4$3x² + y² = 4, the $3x^2$3x² gives 6x, so $\frac{dy}{dx}=-\frac{3x}{y}$dy/dx = −3x/y, not $-\frac{x}{y}$−x/y.

Turning the answer upside down. From $2y\frac{dy}{dx}=-2x$2y dy/dx = −2x, $\frac{dy}{dx}=-\frac{x}{y}$dy/dx = −x/y, with y in the denominator.

A leaf-shaped panel

In the application below, the edge of a panel follows $x^3+y^3=9xy$x³ + y³ = 9xy, which has a term in y alone and a term with both letters, so it uses the chain rule and the product rule together.