Implicit differentiation

Curves that are not functions

The circle x^2+y^2=25 is not the graph of a function, because most vertical lines cross it twice. Yet at almost every point it has a perfectly good tangent line. How do you find its slope without solving for y?

Drag the ochre point around each curve below and watch the tangent. Where is it horizontal? Where does it stand straight up, so that no slope exists?

Drag the ochre point. It stays on the curve.

Slope formula from implicit differentiation
Curve
x
y
dy/dx
Tangent

The method

Near most points, a piece of the curve is the graph of some function y(x), even if you cannot write it down. So differentiate both sides of the equation with respect to x, treating y as a function of x. Every time you differentiate an expression in y, the chain rule attaches a factor of \frac{dy}{dx}:

\frac{d}{dx}\,y^2=2y\,\frac{dy}{dx},\qquad \frac{d}{dx}\,xy=y+x\,\frac{dy}{dx}.

For the circle, this gives 2x+2y\frac{dy}{dx}=0, so \frac{dy}{dx}=-\frac{x}{y}. Check it at (3,4): the slope should be -\tfrac34. What happens to this formula at (5,0), and what does the graph show there?

The slope formula depends on both x and y. That is expected: at x=3 the circle passes through (3,4) and (3,-4), with opposite slopes.

Solve it step by step

Differentiate both sides, collect the dy/dx terms, and solve. Decimals to two places are fine. The score counts steps you answer correctly on the first try.

Problems solved0
First-try steps0 of 0

    Common mistakes

    Dropping dy/dx

    The derivative of y^2 with respect to x is 2y\frac{dy}{dx}, not 2y. Without the chain-rule factor there is nothing to solve for.

    Forgetting the product rule

    A term like xy has two factors that both depend on x. Its derivative is y+x\frac{dy}{dx}.

    Differentiating a constant as if it changes

    The right side of x^2+y^2=25 differentiates to 0.