Implicit Differentiation — AP® Calculus AB Notes

So far every function came as $y =$ a formula in $x$ . Many curves do not. The circle $x^2 + y^2 = 25$ and the tilted oval $x^2 + xy + y^2 = 7$ mix $x$ and $y$ in one equation, and solving for $y$ is either messy or gives two formulas. Implicit differentiation finds the slope at a point straight from the equation, with no solving for $y$ first.

Explicit and implicit

A function is given explicitly when $y$ stands alone: $y = \sqrt{25 - x^2}$ . It is given implicitly when an equation links $x$ and $y$ , such as $x^2 + y^2 = 25$ . The whole circle fails the vertical line test, so it is not the graph of one function. But zoom in near the point $(3, 4)$ and you see a small arc that does pass the test. On that arc, $y$ is a function of $x$ , even if we never write its formula, and its slope is what we want.

The tilted ellipse x² + xy + y² = 7 in crimson on axes from −4 to 4. A dashed vertical line x = 1 meets it at (1, 2) and (1, −3), so y is not a function of x. A blue tangent line at (1, 2) has slope −4/5. — **Figure 1.** The curve $x^2 + xy + y^2 = 7$ . The line $x = 1$ meets it twice, yet near $(1, 2)$ the curve has one tangent line, with slope $-\frac{4}{5}$ . LevlPrep original diagram.

The key step: the chain rule on $y$

Treat $y$ as an unknown function of $x$ . Then a term like $y^2$ is a composite function: the inside is $y$ , the outside is the square. By the chain rule,

$\frac{d}{dx}\left(y^2\right) = 2y \cdot \frac{dy}{dx}$

The factor $\frac{dy}{dx}$ is the derivative of the inside function. It appears every time you differentiate a term that holds $y$ , and never for a term that holds only $x$ :

Term	Derivative with respect to $x$	Why
$x^3$	$3x^2$	only $x$ : no extra factor
$y^3$	$3y^2\frac{dy}{dx}$	chain rule, inside $y$
$\sin y$	$\cos y \cdot \frac{dy}{dx}$	chain rule, inside $y$
$e^y$	$e^y\frac{dy}{dx}$	chain rule, inside $y$
$xy$	$y + x\frac{dy}{dx}$	product rule; the $y$ factor gives $\frac{dy}{dx}$
$7$	$0$	a constant

The routine

Apply $\frac{d}{dx}$ to both sides of the equation and write the new equation in full, with its right side (often $= 0$ ).
Move every term that holds $\frac{dy}{dx}$ to one side and everything else to the other.
Factor out $\frac{dy}{dx}$ and divide.
For a slope at a point, substitute both coordinates after differentiating.

Worked example 1. Find the slope of the circle $x^2 + y^2 = 25$ at $(3, 4)$ .

Differentiate both sides: $2x + 2y\frac{dy}{dx} = 0$ .

Solve: $\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}$ .

At $(3, 4)$ : $\frac{dy}{dx} = -\frac{3}{4}$ . Check with the picture: at the upper right of the circle the curve falls as $x$ grows, so a negative slope makes sense.

Worked example 2. Find $\frac{dy}{dx}$ for $x^2 + xy + y^2 = 7$ , and the slope at $(1, 2)$ .

Check the point first: $1 + 2 + 4 = 7$ , so $(1, 2)$ is on the curve.

Differentiate both sides, using the product rule on $xy$ : $2x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx}$ $= 0$ .

Collect: $(x + 2y)\frac{dy}{dx}$ $= -(2x + y)$ , so $\frac{dy}{dx}$ $= -\frac{2x + y}{x + 2y}$ .

At $(1, 2)$ : $\frac{dy}{dx}$ $= -\frac{2 + 2}{1 + 4}$ $= -\frac{4}{5}$ . The tangent line is $y - 2 = -\frac{4}{5}(x - 1)$ .

Worked example 3. Find $\frac{dy}{dx}$ for $e^y + y = x^2$ .

Differentiate: $e^y\frac{dy}{dx} + \frac{dy}{dx} = 2x$ . Both terms on the left hold $y$ , so both get $\frac{dy}{dx}$ .

Factor: $(e^y + 1)\frac{dy}{dx} = 2x$ , so $\frac{dy}{dx} = \frac{2x}{e^y + 1}$ .

Reading the answer

The formula for $\frac{dy}{dx}$ usually holds both $x$ and $y$ . That is not a flaw: on the circle, the points $(3, 4)$ and $(3, -4)$ share $x = 3$ but have slopes $-\frac{3}{4}$ and $\frac{3}{4}$ . You need both coordinates to know which point you mean, and the point must lie on the curve.

A fraction for $\frac{dy}{dx}$ also shows special points. Where the numerator is 0 and the denominator is not, the slope is 0 and the tangent line is flat. Where the denominator is 0 and the numerator is not, the slope is undefined and the tangent line is vertical, as at $(-5, 0)$ and $(5, 0)$ on the circle. Always check that such a point is actually on the curve.

Slips that cost points

Forgetting $\frac{dy}{dx}$ on a $y$ -term. Writing $2x + 2y = 0$ for the circle loses the whole problem.
Skipping the product rule. $\frac{d}{dx}(xy)$ is $y + x\frac{dy}{dx}$ , two terms, not $\frac{dy}{dx}$ alone.
Differentiating one side only. Apply $\frac{d}{dx}$ to both sides, and keep the " $= 0$ ".
Substituting too early. Put in the point's coordinates only after differentiating; a number has derivative 0.
Mixing notations. Keep $\frac{dy}{dx}$ (or $y'$ ) separate from $dy$ and from $\frac{dy}{dt}$ .

Explicit and implicit

The key step: the chain rule on yy

The routine

Reading the answer

Slips that cost points

The key step: the chain rule on $y$