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FFR - Implicit Differentiation

This original textbook reading covers part of the AP topic. Follow the AP course outline for complete topic instruction and practice.

This reading includes verified answer corrections and explicitly stated clarifications.

Warning: Implicit language

We have so far only been learning to take explicit derivatives, y in terms of x. However, it is possible to take derivatives of functions where y is implied within the function. Often times, this is actually easier than taking the derivative through conventional means. For example, how would we differentiate the equation $y = 3 x y + x^{2}$ ? Conventionally, we might subtract both sides by 3xy, factor out a y, and divide by 1-3x on both sides to get y by itself. The final result would look something like this:
$y = \frac{x^{2}}{1 - 3 x}$ You can imagine that finding this function’s derivative would be pretty tedious, involving the use of the quotient rule.

What about equations like $y = x \cdot \sin (y)$ ? Differentiating that would be hopeless, even with the quotient and product rules. However, the chain rule lets us do something special with that function, known as implicit differentiation.

Since y is just some function of x, we can use the chain rule to write its derivative in a slightly different way. Suppose we have an equation, y= 3y² + x². We can take the derivative with respect to x on both sides of the equation to get:

$\frac{d y}{d x} = \frac{d}{d x} (3 y^{2} + x^{2})$

We run into a problem here though. Since y is a function of x, $\frac{d}{d x} 3 y^{2} \neq 6 y$ !! That would be ignoring everything we have learned from the chain rule, which tells us that the derivative of that term would be equal to the derivative of the outside function, 3y², times the derivative of the inside function, y. What’s the derivative of y with respect to x? It’s just $\frac{d y}{d x}$ ! So using the chain rule we see that

$\frac{d y}{d x} = \frac{d}{d x} (3 y^{2} + x^{2}) = 6 y \cdot \frac{d y}{d x} + 2 x$

Another way of thinking about this is

$\frac{d y}{d x} = \frac{d}{d x} 3 y^{2} + \frac{d}{d x} x^{2} = \frac{d}{d y} 3 y^{2} \cdot \frac{d y}{d x} + \frac{d}{d x} x^{2}$

Notice that the dy’s “cancel out” so that we are still taking the derivative with respect to x, except now we can also take the derivative of 3y² with respect to y without breaking the laws of mathematics. Give it a while to sink in, and it will make sense. Let’s try some practice problems in the meantime.

Example Problem 4.4.1

Take the derivative of $y = 3 x y + x^{2}$ using:

conventional methods
using implicit differentiation.

a) Let’s start by isolating y to one side so we can set x in terms of y, making the function explicit. We can accomplish this by first subtracting 3xy on both sides.

$y - 3 x y = x^{2}$

$y \cdot (1 - 3 x) = x^{2}$

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

Now we can take the derivative using the quotient rule.

$y^{'} = \frac{((1 - 3 x) \cdot \frac{d}{d x} x^{2}) - (x^{2} \cdot \frac{d}{d x} (1 - 3 x))}{(1 - 3 x)^{2}}$ $y^{'} = \frac{(1 - 3 x) \cdot 2 x - x^{2} \cdot (- 3)}{(1 - 3 x)^{2}}$

$y' = \frac{2 x - 3 x^{2}}{{(1 - 3 x)}^{2}}$ . The numerator is 2x(1 − 3x) + 3x² = 2x − 3x².

b) We don’t need to set y in terms of x for part b. We can just differentiate implicitly using the chain rule.

$y = x^{2} + 3 x y$
$\frac{d y}{d x} = 2 x + \frac{d}{d x} 3 x y$

using the product rule, we get

$\frac{d y}{d x} = 2 x + 3 y + 3 x \cdot 1 \cdot \frac{d y}{d x}$
Here, we can isolate the derivative terms to one side to find out what the derivative function is.
$\frac{d y}{d x} - 3 x \cdot \frac{d y}{d x} = 2 x + 3 y$
$\frac{d y}{d x} (1 - 3 x) = 2 x + 3 y$

$\frac{d y}{d x} = \frac{2 x + 3 y}{1 - 3 x}$

Normally this would be enough, but we want to compare our answer to part a, so let’s substitute y because we set it in terms of x explicitly earlier.

Substituting $y = \frac{x^{2}}{1 - 3 x}$ gives:

$\frac{d y}{d x} (1 - 3 x) = 2 x + \frac{3 x^{2}}{1 - 3 x}$
Let’s find a common denominator for the two terms on the right hand side before dividing through by (1-3x)

$\frac{d y}{d x} (1 - 3 x) = 2 x \cdot (\frac{1 - 3 x}{1 - 3 x}) + \frac{3 x^{2}}{1 - 3 x}$
$\frac{d y}{d x} (1 - 3 x) = \frac{2 x - 6 x^{2} + 3 x^{2}}{1 - 3 x}$

Now we can divide by (1-3x) on both sides and we find that

$\frac{d y}{d x} = \frac{2 x - 3 x^{2}}{(1 - 3 x)^{2}}$ We get the same answer! Amazing!

You might be thinking: “My dear sir, would it not have been simpler to just take the derivative using the quotient rule after isolating y on one side?” And you would have a great point. Implicit differentiation is simply another tool in our arsenal that can help us solve derivatives. The point of this example problem was to show that it is one and the same thing as setting y explicitly in terms of x and then differentiating. Besides, would the quotient or product rule help you solve the next problem?

Example Problem 4.4.2

Find the derivative of $y^{2} = \sin x \cos y + x^{4}$

We have no choice but to use implicit differentiation here because there is no simple way to isolate y. Remember that whenever we take the derivative of a term with “y” in it, we have to use the chain rule and multiply by $\frac{d y}{d x}$ after we finish taking the derivative.

$2 y \cdot \frac{d y}{d x} = \frac{d}{d x} (\sin x \cos y) + 4 x^{3}$
$2 y \cdot \frac{d y}{d x} = \cos x \cos y + \sin x \cdot (- \sin y \cdot \frac{d y}{d x}) + 4 x^{3}$
Let’s get the dy/dx terms on one side. Notice the two sine functions have different arguments, so we cannot combine them.

$2 y \cdot \frac{d y}{d x} + \sin x \sin y \cdot \frac{d y}{d x} = \cos x \cos y + 4 x^{3}$

$\frac{d y}{d x} \cdot (2 y + \sin x \sin y) = \cos x \cos y + 4 x^{3}$

Now we can solve for the derivative:

$\frac{d y}{d x} = \frac{\cos x \cos y + 4 x^{3}}{2 y + \sin x \sin y}$

And this gives us our final answer, but there’s a conundrum! How would we graph this function, if it even is a function? We could try going back and isolating y, but that would be too tedious and not worth the effort. We could try something else though. Let’s create a three-dimensional graph. It’ll be just like real life! Refer to Gallery 4.4.1.

Gallery 4.4.1 Graph of example 4.4.2

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Image 1

Despite what your intuition might tell you, the y-axis is the one in the back. The z-axis (representing dy/dx) is the one going up and down. As usual, the x-axis goes left and right. It takes a while getting accustomed to reading 3-D graphs so take a look at this one carefully.

Image 2

This view of the graph shows the x-axis pointing directly at us. It allows us to look at the y-z plane. Notice the function is undefined when y= 0. The curvy contour lines that we see on the surface of the graphs have a special name, a “family of functions” that differ only by a constant.

Image 3

This view of the graph shows the y-axis pointing away from us, which conveniently allows us to see the change in the derivative function with respect to the x-value. These graphs look really neat and show us how the derivative behaves, but do they tell the whole story?

Image 4

Hopefully some of you noticed that we were missing something while being distracted by the pretty 3-D graphs. The derivative function has no restrictions the way it is written, so it doesn’t tell the whole story. What do you think this new graph represents?

Image 5

Correct! It is our parent function, y²=sin(y)cos(y)+x⁴. This is the function viewed on the x-y plane, how we are probably used to looking at it in two dimensions. Notice that since it has no z-component, it extends infinitely in the z-drection because it is independent of its derivative.

Image 6

From this angle, we can see what the intersection of the two graphs looks like. The curves where the two graphs intersect represent the fulfillment of two conditions: 1) Their x and y coordinates represent actual solutions to our base function, and

Image 7

2) Their derivative at these points is represented by the z-coordinate.
This view of the graph shows the x-axis moving away from us. You can compare the difference with the previous graph, which shows positive x-values. This curve represents negative x-values.

Show your work for Problem 1

a) Implicitly differentiate the equation

y = x \cdot y \cdot e^{x} + 1

b) Write its derivative as a function of just x.

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Chapter 4 Problems

a) Implicitly differentiate the equation $y = x \cdot y \cdot e^{x} + 1$
b) Write its derivative as a function of just x.
Differentiate

Show your work for Problem 2

Differentiate

f (x) = \frac{\tan x}{x^{2} - 4}

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$f (x) = \frac{\tan x}{x^{2} - 4}$

Show your work for Problem 3

Find the derivative of

y = \frac{e^{x} - 4 x}{\sin^{2} x}

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Find the derivative of $y = \frac{e^{x} - 4 x}{\sin^{2} x}$
Using the equation $\frac{(x - 4)^{2}}{25} + \frac{(y + 1)^{2}}{16} = 1$ , find the slope of the tangent line at x=1 in the first quadrant

Show your work for Problem 4

Using the equation

\frac{(x - 4)^{2}}{25} + \frac{(y + 1)^{2}}{16} = 1

, find the slope of the tangent line at x=1 in the first quadrant .

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Show your work for Problem 5

Differentiate

y = \frac{x y + 3 e^{x}}{y^{2}}

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Show your work for Problem 6

Differentiate

y = \frac{x \sin y + 2 \cos (2 x^{2})}{x^{3}}

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Differentiate $y = \frac{x y + 3 e^{x}}{y^{2}}$
Differentiate $y = \frac{x \sin y + 2 \cos (2 x^{2})}{x^{3}}$

Show your work for Problem 7

Differentiate

y = x^{2} \cdot \sin y \cdot e^{\cos x}

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Show your work for Problem 8

Using the equation given below, find the slope of its tangent line at

x = \frac{π}{4}

(Ref: Unit Circle):

\begin{aligned} y^{2} + 1 & = y \cdot \sin x + y \cdot \cos (2 x) + \frac{\sin x}{y} \\ + \frac{\cos^{2} x}{y} - \frac{\sin^{2} x}{y} \end{aligned}

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Differentiate $y = x^{2} \cdot \sin y \cdot e^{\cos x}$
Using the equation given below, find the slope of its tangent line at $x = \frac{π}{4}$ (Ref: Unit Circle): $\begin{aligned} y^{2} + 1 & = y \cdot \sin x + y \cdot \cos (2 x) + \frac{\sin x}{y} \\ + \frac{\cos^{2} x}{y} - \frac{\sin^{2} x}{y} \end{aligned}$

Show your work for Problem 9

Find the implicit derivative of

\frac{x y + x^{2}}{e^{2 x}} = \frac{\cos (4 x^{2} + 1) \sin y}{e^{x}}

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Show your work for Problem 10

Find the implicit derivative of

y = x \sin (\ln (y)) + \frac{x y}{\cos x}

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Find the implicit derivative of $\frac{x y + x^{2}}{e^{2 x}} = \frac{\cos (4 x^{2} + 1) \sin y}{e^{x}}$
Find the implicit derivative of $y = x \sin (\ln (y)) + \frac{x y}{\cos x}$

Glossary and deeper reading · 3 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

Derivative
A derivative gives the instantaneous rate of change of a function with respect to its input. At a point, it is the finite limit of the difference quotient when that limit exists. For example, the derivative of x² is 2x, so the tangent slope at x=3 is 6. A derivative function collects those rates across the inputs where the function is differentiable.
Read more: 2.2 · Derivative →
Function
A mathematical function assigns exactly one output to each input in its domain. Inputs can share an output, but one input cannot have two different outputs. For example, f(x)=x² assigns the output 9 to both 3 and −3. Function notation names the rule and its input; f(3) means the output when the input is 3.
Chain rule
The chain rule differentiates a composition by multiplying the outer derivative at the inner output by the inner derivative. For differentiable functions, (f(g(x)))′=f′(g(x))g′(x). For example, the derivative of sin(x²) is cos(x²)·2x. Nested compositions require a factor for each successive layer.
Read more: 3.1 · Chain rule →

Interactive 4.4.1 3-D graph for problem 4.4.2

Gallery 4.4.1 Graph of example 4.4.2

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Show your work for Problem 1

Show your work for Problem 2

Show your work for Problem 3

Show your work for Problem 4

Show your work for Problem 5

Show your work for Problem 6

Show your work for Problem 7

Show your work for Problem 8

Show your work for Problem 9

Show your work for Problem 10

Derivative

Function

Chain rule

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