What is the difference between implicit and explicit differentiation?

Explicit differentiation starts from a formula y=f(x). Implicit differentiation starts from an equation relating x and y, differentiates both sides, and solves for (dy)/(dx) afterward.

Why do you multiply by dy/dx when differentiating y?

Because y is a function of x, so an expression like y^3 is a composite function. The chain rule multiplies the outside's derivative by the inside's derivative, which is (dy)/(dx).

Can the answer to implicit differentiation contain y?

Yes, and it usually does. To get a numerical slope, substitute the coordinates of a point on the curve.

Do you have to solve for y first?

No. That's the point of the method: it works even when solving for y is impossible, as for x^3+y^3=6xy.

How do you differentiate xy implicitly?

With the product rule: (d)/(dx)(xy)=1· y+x(dy)/(dx).

Implicit Differentiation: Step-by-Step Method and Examples

When do you need implicit differentiation?

Most functions come as a formula, $y =$ something in $x$ , and the rules differentiate it directly. But some curves are given by an equation relating $x$ and $y$ , such as the circle $x^{2} + y^{2} = 25$ . Solving for $y$ is either awkward (here it gives two formulas, $y = \pm \sqrt{25 - x^{2}}$ ) or impossible, as for $x^{3} + y^{3} = 6 x y$ .

An equation like this still describes $y$ as depending on $x$ near most points; it just doesn't say how in a formula. We say the equation defines $y$ implicitly (by implication), as opposed to explicitly (spelled out as $y = \dots$ ). Implicit differentiation finds the slope $\frac{d y}{d x}$ without ever solving for $y$ .

How does implicit differentiation work?

Think of $y$ as a hidden function of $x$ , say $y = y (x)$ . Then $y^{2}$ is really $(y (x))^{2}$ : a square with $y (x)$ inside. The chain rule says to differentiate the outside (giving $2 y$ ) and multiply by the inside's derivative (which is $\frac{d y}{d x}$ ). The same happens to every expression in $y$ :

Expression	Derivative with respect to $x$
$y$	$\frac{d y}{d x}$
$y^{2}$	$2 y \frac{d y}{d x}$
$y^{3}$	$3 y^{2} \frac{d y}{d x}$
$\sin y$	$\cos y \frac{d y}{d x}$
$e^{y}$	$e^{y} \frac{d y}{d x}$
$x y$ (a product)	$1 \cdot y + x \frac{d y}{d x}$

Terms in $x$ alone are differentiated as usual, and constants give $0$ .

What are the steps of implicit differentiation?

Differentiate both sides of the equation with respect to $x$ .
Wherever you differentiate an expression in $y$ , multiply by $\frac{d y}{d x}$ (chain rule). Use the product rule for terms like $x y$ .
Collect all the terms containing $\frac{d y}{d x}$ on one side and everything else on the other.
Factor out $\frac{d y}{d x}$ and divide to solve for it.

Worked example: A circle

Find $\frac{d y}{d x}$ for $x^{2} + y^{2} = 25$ , and the slope at the point $(3, 4)$ .

Differentiate both sides: $\frac{d}{d x} (x^{2}) + \frac{d}{d x} (y^{2}) =$ $\frac{d}{d x} (25)$ .
This gives $2 x + 2 y \frac{d y}{d x} =$ $0$ ; the right side is $0$ because $25$ is a constant.
Move $2 x$ to the right: $2 y \frac{d y}{d x} =$ $- 2 x$ .
Divide by $2 y$ : $\frac{d y}{d x} =$ $- \frac{x}{y}$ .
At $(3, 4)$ , which is on the circle since $9 + 16 = 25$ : $\frac{d y}{d x} =$ $- \frac{3}{4}$ .

Answer: $\frac{d y}{d x} =$ $- \frac{x}{y}$ , and the slope at $(3, 4)$ is $- \frac{3}{4}$ .

The circle

x^{2} + y^{2} = 25

and its tangent line at

(3, 4)

, with slope

- \frac{3}{4}

The answer depends on both $x$ and $y$ , which makes sense: at $x = 3$ the circle has two points, $(3, 4)$ and $(3, - 4)$ , with different slopes ( $- \frac{3}{4}$ and $\frac{3}{4}$ ). The coordinates of the point tell the formula which one you mean.

Worked example: A product term

Find $\frac{d y}{d x}$ for $x^{2} + x y + y^{2} = 7$ , and the tangent line at $(1, 2)$ .

Check the point: $1 + 2 + 4 = 7$ . ✓
Differentiate term by term. $x^{2} \to 2 x$ . The product $x y \to 1 \cdot y + x \frac{d y}{d x}$ . $y^{2} \to 2 y \frac{d y}{d x}$ . $7 \to 0$ .
So $2 x + y + x \frac{d y}{d x} + 2 y \frac{d y}{d x} =$ $0$ .
Collect the $\frac{d y}{d x}$ terms: $(x + 2 y) \frac{d y}{d x} =$ $- (2 x + y)$ .
Divide: $\frac{d y}{d x} =$ $- \frac{2 x + y}{x + 2 y}$ .
At $(1, 2)$ : $- \frac{2 + 2}{1 + 4} =$ $- \frac{4}{5}$ . The tangent line is $y - 2 = - \frac{4}{5} (x - 1)$ , which simplifies to $y =$ $- \frac{4}{5} x + \frac{14}{5}$ .

Answer: $\frac{d y}{d x} =$ $- \frac{2 x + y}{x + 2 y}$ ; tangent line $y =$ $- \frac{4}{5} x + \frac{14}{5}$ .

Common mistake

Forgetting $\frac{d y}{d x}$ on the $y$ -terms. Writing $\frac{d}{d x} y^{2} = 2 y$ is the most common mistake: $y$ depends on $x$ , so the chain rule adds the factor $\frac{d y}{d x}$ . A second common slip is treating $x y$ as if it had derivative $1 \cdot \frac{d y}{d x}$ instead of using the product rule.

How do you find horizontal and vertical tangent lines implicitly?

Write $\frac{d y}{d x}$ as a fraction. A horizontal tangent has slope $0$ , so the top of the fraction is $0$ (and the bottom isn't). A vertical tangent has no slope, which happens where the bottom is $0$ (and the top isn't). In both cases, use the original equation to find the points.

Worked example: Horizontal and vertical tangents of a circle

Where does $x^{2} + y^{2} = 25$ have horizontal and vertical tangent lines?

From the first example, $\frac{d y}{d x} =$ $- \frac{x}{y}$ .
Horizontal: top $= 0$ , so $x = 0$ . Then $y^{2} = 25$ , so $y = \pm 5$ : the points $(0, 5)$ and $(0, - 5)$ .
Vertical: bottom $= 0$ , so $y = 0$ . Then $x^{2} = 25$ : the points $(5, 0)$ and $(- 5, 0)$ .

Answer: Horizontal at $(0, \pm 5)$ , vertical at $(\pm 5, 0)$ : the top, bottom and sides of the circle.

Where does the derivative of ln x come from?

Implicit differentiation also explains several formulas. If $y = \ln x$ , then by the meaning of the natural logarithm, $e^{y} = x$ . Differentiate both sides: $e^{y} \frac{d y}{d x} = 1$ , so $\frac{d y}{d x} =$ $\frac{1}{e^{y}}$ . Since $e^{y} = x$ , this is $\frac{1}{x}$ . The same idea gives the derivatives of the inverse trig functions, as covered in derivatives of $e^{x}$ and $\ln x$ .

How do you find the second derivative implicitly?

Differentiate the formula for $\frac{d y}{d x}$ again (usually with the quotient rule), remembering that $y$ still depends on $x$ , then substitute the first derivative and simplify using the original equation.

Worked example: Second derivative of a circle

Find $\frac{d^{2} y}{d x^{2}}$ for $x^{2} + y^{2} = 25$ .

Start from $y^{'} = - \frac{x}{y}$ , writing $y^{'}$ for $\frac{d y}{d x}$ .
Quotient rule, with $\frac{d}{d x} y = y^{'}$ : $y^{″} =$ $- \frac{1 \cdot y - x y^{'}}{y^{2}}$ .
Substitute $y^{'} = - \frac{x}{y}$ : the top is $y - x (- \frac{x}{y}) =$ $y + \frac{x^{2}}{y} =$ $\frac{y^{2} + x^{2}}{y}$ .
So $y^{″} =$ $- \frac{y^{2} + x^{2}}{y^{3}}$ , and since $x^{2} + y^{2} = 25$ : $y^{″} =$ $- \frac{25}{y^{3}}$ .

Answer: $\frac{d^{2} y}{d x^{2}} =$ $- \frac{25}{y^{3}}$

Tip

Implicit differentiation is the engine behind related rates: there, you differentiate an equation with respect to time $t$ , so every changing quantity gets its own rate factor.

Implicit differentiation, step by step