AP Calculus AB Implicit Differentiation Lesson (Free PDF)

AP Calculus AB math lesson: Implicit Differentiation – Maths PDF

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This implicit lesson for AP Calculus AB covers implicit differentiation (Common Core AP CALC AB 3.2), with a free PDF to print.

Implicit differentiation finds dy/dx for an equation relating x and y without solving for y. Differentiate both sides with respect to x, multiply by dy/dx every time you differentiate a y term (chain rule), then solve for dy/dx. For x² + y² = 25, the result is dy/dx = -x/y.

AP Calculus AB · Common Core State Standards: AP CALC AB 3.2

Key facts in this implicit lesson

Key facts

Differentiate every term with respect to x; a y term gets a factor dy/dx.
Use the product rule for terms such as xy, and the chain rule for sin y or e^y.
The derivative may contain both x and y; substitute a point on the curve to get a slope.
Horizontal tangents: numerator zero. Vertical tangents: denominator zero.

What is implicit differentiation?

Some curves, such as the circle \(x^{2}+y^{2}=25\), cannot be written as a single function \(y=f(x)\). The equation defines \(y\) implicitly. Implicit differentiation finds \(\frac{dy}{dx}\) without solving for \(y\) first.

Key idea

Treat \(y\) as a function of \(x\). Whenever you differentiate a term with \(y\), use the chain rule: \(\frac{d}{dx}[y^{n}]=ny^{n-1}\frac{dy}{dx}\). Terms with only \(x\) are differentiated as usual.

Four steps for implicit differentiation.

How do you do it step by step?

Example. Find \(\frac{dy}{dx}\) for \(x^{2}+y^{2}=25\).

Differentiate both sides: \(2x+2y\frac{dy}{dx}=0\).
Isolate the \(\frac{dy}{dx}\) term: \(2y\frac{dy}{dx}=-2x\).
Solve: \(\frac{dy}{dx}=-\frac{x}{y}\).

At the point \((3,4)\), the slope is \(-\frac{3}{4}\), so the tangent line is \(y-4=-\frac{3}{4}(x-3)\).

At (3, 4) the slope of the circle is −x/y = −3/4.

How do products and other terms work?

For \(xy\), use the product rule: \(\frac{d}{dx}[xy]=y+x\frac{dy}{dx}\). For \(\sin y\), use the chain rule: \(\cos y\cdot\frac{dy}{dx}\). For \(e^{y}\), the result is \(e^{y}\frac{dy}{dx}\).

When the derivative has \(y\) in it, that is normal. Substitute both coordinates of a point to get a numerical slope. The point must satisfy the original equation.

Example with a product. For \(xy=6\), differentiate both sides: \(y+x\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac{y}{x}\). At \((2,3)\) the slope is \(-\frac{3}{2}\). You can confirm this by solving \(y=\frac{6}{x}\) and using the power rule: \(y^{\prime}=-\frac{6}{x^{2}}=-\frac{6}{4}=-\frac{3}{2}\).

The curve xy = 6 with its tangent at (2, 3); slope −3/2.

When are tangents horizontal or vertical?

A tangent is horizontal where the numerator of \(\frac{dy}{dx}\) is zero and the denominator is not. It is vertical where the denominator is zero and the numerator is not. For \(9x^{2}+4y^{2}=36\), \(\frac{dy}{dx}=-\frac{9x}{4y}\), so tangents are horizontal at \((0,\pm 3)\) and vertical at \((\pm 2,0)\).

Horizontal tangents where dy/dx = 0; vertical tangents where the denominator is 0.

A common check is to pick a point, compute the slope with implicit differentiation, and compare it with the picture. On the circle, the slope is negative in the first quadrant and positive in the second quadrant, exactly as the graph shows.

For a second derivative, differentiate \(\frac{dy}{dx}\) again, and replace \(\frac{dy}{dx}\) with its formula. For the circle, \(\frac{d^{2}y}{dx^{2}}=-\frac{25}{y^{3}}\).

Implicit lesson: questions and answers

What is implicit differentiation used for?

It finds slopes of curves that are not written as y = f(x), such as circles and ellipses, and it handles equations that are hard or impossible to solve for y.

Why do you multiply by dy/dx?

Because y is a function of x, so differentiating a y expression requires the chain rule. For example, the derivative of y squared with respect to x is 2y times dy/dx.

How do you find the slope at a point on an implicit curve?

Find dy/dx in terms of x and y, then substitute both coordinates of the point. Check first that the point satisfies the original equation.

How do you find a vertical tangent?

Write dy/dx as a fraction and find where the denominator is zero while the numerator is nonzero. Then use the original equation to find the matching points.

How do you find a second derivative implicitly?

Differentiate dy/dx with respect to x, apply the chain rule to every y, and then substitute the expression for dy/dx. Simplify using the original equation if possible.

How to use this implicit lesson

Read the definitions first, then follow the worked method line by line and redo the example on paper without looking. When every step makes sense, move on to the worksheet, then check yourself with the quiz.

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