Implicit Differentiation, calculus lesson

A circle is easy to draw but it is not the graph of a single function, since most vertical lines meet it twice. Still, it has a tangent line at every point, and away from the far left and right points that tangent has an ordinary finite slope. Implicit differentiation finds the slope of that tangent straight from the equation, even for curves where solving for $y$ is hard or impossible.

Key terms

Implicit curves

Definition An implicit curve is the set of points $(x, y)$ in the plane that satisfy an equation $F(x, y) = 0$ , or more generally an equation between two expressions in $x$ and $y$ . The equation defines the curve implicitly, as opposed to an explicit formula $y = f(x)$ .

In plain words The curve is everything that makes the equation true. It may loop, cross itself, or have several points above the same $x$ .

Example The circle $x^2 + y^2 = 25$ contains both $(3, 4)$ and $(3, -4)$ , so it is not the graph of a function of $x$ . The curve $x^3 + y^3 = 6xy$ passes through $(3, 3)$ , since $27 + 27 = 54 = 6 \cdot 3 \cdot 3$ .

Local branches

Definition Let $(a, b)$ be a point of the curve. A local branch of the curve through $(a, b)$ is a function $y = g(x)$ with $g(a) = b$ , defined on an open interval $I$ containing $a$ , such that for some open interval $J$ containing $b$ the points of the curve inside the rectangle $I \times J$ are exactly the points $(x, g(x))$ with $x$ in $I$ . For a curve $F(x, y) = 0$ , the implicit function theorem from multivariable calculus says that such a branch exists, and is differentiable, whenever $F$ has continuous first partial derivatives near $(a, b)$ and $\frac{\partial F}{\partial y}(a, b) \neq 0$ .

In plain words Zoom in close enough to a point, and the curve usually looks like the graph of an ordinary function. That small piece is a branch, and it is what $\frac{dy}{dx}$ describes.

Example On the circle $x^2 + y^2 = 25$ , the branch through $(3, 4)$ is $y = \sqrt{25 - x^2}$ for $-5 < x < 5$ , with $J = (0, 6)$ for example, and the branch through $(3, -4)$ is $y = -\sqrt{25 - x^2}$ . At $(5, 0)$ there is no branch of the form $y = g(x)$ , because every rectangle around that point holds two points of the circle above each $x$ slightly less than $5$ and none to its right.

Differentiating an equation

Definition Suppose $y = g(x)$ is a differentiable local branch of the curve. Substituting it into the equation gives an identity in $x$ on an open interval, so both sides have equal derivatives there. Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ and using the chain rule, for example

\frac{d}{dx} y^n = n y^{n-1} \frac{dy}{dx}, \qquad \frac{d}{dx} (xy) = y + x \frac{dy}{dx}.

In plain words Differentiate every term as usual, and each time you differentiate something containing $y$ , attach a factor $\frac{dy}{dx}$ .

Example Differentiating $x^2 + y^2 = 25$ with respect to $x$ gives

2x + 2y \frac{dy}{dx} = 0.

The derivative dy/dx

Definition After differentiating the equation, solve the result for $\frac{dy}{dx}$ . The formula obtained gives the slope of the local branch at any point of the curve where a differentiable branch exists and the coefficient of $\frac{dy}{dx}$ is not zero.

In plain words The slope comes out as an expression in both $x$ and $y$ . You need the full point $(x, y)$ to evaluate it, which is how it tells the two branches of a circle apart.

Example Solving $2x + 2y \frac{dy}{dx} = 0$ on the circle gives

\frac{dy}{dx} = -\frac{x}{y}, \qquad y \neq 0.

At $(3, 4)$ the slope is $-\frac{3}{4}$ and at $(3, -4)$ it is $\frac{3}{4}$ . For $x^3 + y^3 = 6xy$ , differentiating gives

3x^2 + 3y^2 y' = 6y + 6xy',

so collecting the $y'$ terms and dividing by $3(y^2 - 2x)$ gives

y' = \frac{2y - x^2}{y^2 - 2x}, \qquad y^2 \neq 2x.

At $(3, 3)$ this is $\frac{6 - 9}{9 - 6} = -1$ , and the tangent line is $y = 3 - (x - 3)$ , or $y = 6 - x$ .

Higher implicit derivatives

Definition To find $\frac{d^2 y}{dx^2}$ , differentiate the formula for $\frac{dy}{dx}$ with respect to $x$ , again treating $y$ as a function of $x$ . Then replace every $\frac{dy}{dx}$ by its formula and simplify, using the original equation where it helps. The result holds on the same points as the formula for $\frac{dy}{dx}$ , provided the branch is twice differentiable there.

In plain words Do it again, then substitute the first derivative back in. The original equation often tidies the answer.

Example On $x^2 + y^2 = 25$ with $y \neq 0$ , $y' = -\frac{x}{y}$ . The quotient rule gives

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}.

At $(3, 4)$ , $y'' = -\frac{25}{64}$ . It is negative, as it should be on the upper half of the circle, which bends downward.

Points where the solved formula fails

Definition The formula for $\frac{dy}{dx}$ breaks down at points of the curve where the coefficient of $\frac{dy}{dx}$ is zero. At such a point the curve may have a vertical tangent, a cusp, or a self crossing, or there may be no differentiable branch $y = g(x)$ at all. These points must be examined separately.

In plain words A zero in the denominator is a warning sign. The curve is doing something that a single function of $x$ cannot do smoothly there.

Example On the circle, $y' = -\frac{x}{y}$ fails at $(5, 0)$ and $(-5, 0)$ , where the tangent lines are vertical. On $x^3 + y^3 = 6xy$ , the formula becomes $\frac{0}{0}$ at the origin, where the curve crosses itself with two tangent lines, $y = 0$ and $x = 0$ . No single slope describes that point.

Check yourself

Find the slope of $xy = 6$ at $(2, 3)$ by implicit differentiation.

Show the answer

Differentiating gives $y + x y' = 0$ , so $y' = -\frac{y}{x}$ and the slope at $(2, 3)$ is $-\frac{3}{2}$ .

For $x^2 - y^2 = 1$ , show that $y'' = -\frac{1}{y^3}$ wherever $y \neq 0$ .

Show the answer

Differentiating gives $2x - 2y y' = 0$ , so $y' = \frac{x}{y}$ . The quotient rule and then $x^2 - y^2 = 1$ give

y'' = \frac{y - x y'}{y^2} = \frac{y - \frac{x^2}{y}}{y^2} = \frac{y^2 - x^2}{y^3} = -\frac{1}{y^3}.

Where does the implicit formula for $\frac{dy}{dx}$ fail on the curve $y^2 = x^3$ ?

Show the answer

Differentiating gives $2y y' = 3x^2$ , so $y' = \frac{3x^2}{2y}$ , which fails where $y = 0$ . On the curve that means only the origin, where the curve has a cusp.

Implicit Differentiation

Theory, part 57

Practice, part 58

Key terms

Implicit curves

Local branches

Differentiating an equation

The derivative dy/dx

Higher implicit derivatives

Points where the solved formula fails

Check yourself

How this lesson connects

Builds on

Used later in