D.2 Implicit Differentiation

On the preceding screen we illustrated what an "implicit function" is, including that the term "implicit function" really means "a function defined implicitly." As promised, let's now further develop our Calculus tools to take the derivative of such a function and find $𝑑 𝑦 𝑑 𝑥$ or $𝑦' (𝑥) .$ We'll of course work some examples, and provide practice problems so you can develop your skills.

How to Take an Implicit Derivative

For most students, the process of implicit differentiation is straightforward, building on all of the other "taking derivatives" work we've done. There is one new piece you just have to get used to, which is using the Chain Rule in a particular way. It's in the first bullet point of Step 1 in our Problem Solving Strategy:

Problem Solving Strategy: Implicit Differentiation

There are two basic steps to solve implicit differentiation problems:

Take the derivative $\dfrac{d}{dx}$ of both sides of the equation.
- Use your usual Rules of Differentiation, with one addition: When you take the derivative of a term with a y in it, be sure to multiply by $𝑑 𝑦 𝑑 𝑥$ due to the Chain Rule.
  
  Let's consider, as an example, the function $𝑒 𝑥 2$ . When you take its derivative, you of course use the Chain Rule:
  $𝑑 𝑑 𝑥 (𝑒 𝑥 2) = 𝑒 𝑥 2 \cdot 𝑑 𝑑 𝑥 (𝑥 2) = 𝑒 𝑥 2 \cdot (2 𝑥)$
  Similarly, when you take the derivative of, say, $𝑒 𝑓 (𝑥)$ , where $𝑓 (𝑥)$ is some function we're not specifying, you again of course use the Chain Rule: $𝑑 𝑑 𝑥 (𝑒 𝑓 (𝑥)) = 𝑒 𝑓 (𝑥) \cdot 𝑑 𝑑 𝑥 (𝑓 (𝑥)) = 𝑒 𝑓 (𝑥) \cdot 𝑑 𝑓 𝑑 𝑥$ That might look funny, but we don't know what $𝑑 𝑓 𝑑 𝑥$ is, so we're just leaving it written like that.
  
  And also similarly, let's consider the derivative of, say, $𝑒 𝑦 (𝑥)$ , where we're writing $𝑦 (𝑥)$ to indicate that y is a function of x. We again of course must use the Chain Rule:
  
  $𝑑 𝑑 𝑥 (𝑒 𝑦 (𝑥)) = 𝑒 𝑦 (𝑥) \cdot 𝑑 𝑑 𝑥 (𝑦 (𝑥)) = 𝑒 𝑦 (𝑥) \cdot 𝑑 𝑦 (𝑥) 𝑑 𝑥$ Now we don't usually write $𝑦 (𝑥)$ , because it's annoying to write again and again; instead we just write "y"—but always have to keep in mind that really y still depends on x. We thus usually write that last result as
  
  $𝑑 𝑑 𝑥 (𝑒 𝑦) = 𝑒 𝑦 \cdot 𝑑 𝑦 𝑑 𝑥$
  
  The key point is that we end up multiplying by $𝑑 𝑦 𝑑 𝑥$ , because the function we were taking the derivative of, $𝑒 𝑦$ , has a y in it. Similarly, you should see that the following are all correct derivatives for these example functions:
  
  $𝑑 𝑑 𝑥 [𝑦] = 1 \cdot 𝑑 𝑦 𝑑 𝑥 = 𝑑 𝑦 𝑑 𝑥 𝑑 𝑑 𝑥 [𝑦 5] = 5 𝑦 4 \cdot 𝑑 𝑦 𝑑 𝑥 𝑑 𝑑 𝑥 [s i n (𝑦)] = c o s (𝑦) \cdot 𝑑 𝑦 𝑑 𝑥 𝑑 𝑑 𝑥 [c o s (𝑦 2)] = - s i n (𝑦 2) \cdot 2 𝑦 \cdot 𝑑 𝑦 𝑑 𝑥 𝑑 𝑑 𝑥 [l n (𝑦)] = 1 𝑦 \cdot 𝑑 𝑦 𝑑 𝑥 𝑑 𝑑 𝑥 [l n (𝑦 5)] = 1 𝑦 5 \cdot 5 𝑦 4 \cdot 𝑑 𝑦 𝑑 𝑥 = 5 𝑦 \cdot 𝑑 𝑦 𝑑 𝑥$
  
  The biggest challenge when learning to do Implicit Differentiation problems is to remember to include this $𝑑 𝑦 𝑑 𝑥$ term when you take the derivative of something that has a y in it. As always, practicing is the way to learn, and you'll get good practice problems below.
- Remember that $𝑑 𝑑 𝑥 (c o n s t a n t) = 0$ . (It's a common error to forget that when doing these problems, even among experts.)
Solve for $\dfrac{dy}{dx}$ .
- Collect all terms with $𝑑 𝑦 𝑑 𝑥$ in them on the left side of the equation, all other terms on the right.
- Factor and divide as necessary to solve for $𝑑 𝑦 𝑑 𝑥$ .

Note: You may use $𝑦'$ instead of $𝑑 𝑦 𝑑 𝑥$ . They are interchangeable:

𝑦' = 𝑑 𝑦 𝑑 𝑥

Let's consider an Example to illustrate, returning again to our unit circle.

Example 1: $𝑥 2 + 𝑦 2 = 1$

(a) Given $𝑥 2 + 𝑦 2 = 1,$ find $𝑑 𝑦 𝑑 𝑥 .$

(b) Find the slope of the tangent lines to the circle at $𝑥 = 0.5$ .

Solution.

(a) Let's find $𝑑 𝑦 𝑑 𝑥$ :

Step 1. Take the derivative $𝑑 𝑑 𝑥$ of both sides of the equation. Remember the Chain Rule as applied to y.

𝑑 𝑑 𝑥 [𝑥 2] + 𝑑 𝑑 𝑥 [𝑦 2] = 𝑑 𝑑 𝑥 [1] 2 𝑥 + 2 𝑦 \cdot 𝑑 𝑦 𝑑 𝑥 = 0

Step 2. Solve for

𝑑 𝑦 𝑑 𝑥 .

2 𝑦 \cdot 𝑑 𝑦 𝑑 𝑥 = - 2 𝑥 𝑑 𝑦 𝑑 𝑥 = - 𝑥 𝑦 ✓

Because we can solve the circle equation for y, we aren't required to use implicit differentiation to find $𝑑 𝑦 𝑑 𝑥,$ as illustrated in the box below. However, using implicit differentiation results in a single equation that works for all $(𝑥, 𝑦)$ on the circle, whereas completing the calculation without implicit differentiation requires treating the top and bottom halves of the circle separately, and as you'll see also involves more work.

Furthermore, in situations where we cannot solve the equation for an explicit form of $𝑦 (𝑥) = \dots,$ we have no choice but to use implicit differentiation. (And besides, it's so much easier, there's no reason you would ever want to not use it!)

Insurance against making a mistake

$Tip icon$

The most common error students make, especially on exams, is to forget to use the Chain Rule on one or more terms involving y when taking the derivative. A simple way to help avoid this error is to add a step to the beginning and end of the procedure, and replace y with $𝑓 (𝑥) .$ That is:

Replace y with $𝑓 (𝑥)$ in the equation.
Take the derivative $𝑑 𝑑 𝑥$ of both sides of the equation. Remember the Chain Rule, so every term in the original equation that has an $𝑓 (𝑥)$ will have now contain $𝑑 𝑓 𝑑 𝑥 .$
Solve for $𝑑 𝑓 𝑑 𝑥 .$
If you'd like or if required, substitute back $𝑓 (𝑥) = 𝑦 .$

The next example illustrates. We'll also use prime notation for the derivative to show how that works as well.

Example 2: $𝑥 2 + 𝑦 = 2 𝑦 2 + 1$

Use implicit differentiation to find $𝑦'$ given $𝑥 2 + 𝑦 = 2 𝑦 2 + 1 .$

Solution.

Insurance Process Step 1. Replace y with $𝑓 (𝑥)$ in the equation.

𝑥 2 + 𝑓 (𝑥) = 2 𝑓 (𝑥) 2 + 1

Insurance Process Step 2. Take the derivative of both sides of the equation with respect to x.

2 𝑥 + 𝑓' (𝑥) = 4 𝑓 (𝑥) \cdot 𝑓' (𝑥)

Insurance Process Step 3. Solve for $𝑓' (𝑥) .$

2 𝑥 + 𝑓' (𝑥) = 4 𝑓 (𝑥) \cdot 𝑓' (𝑥) 𝑓' (𝑥) - 4 𝑓 (𝑥) \cdot 𝑓' (𝑥) = - 2 𝑥 𝑓' (𝑥) \cdot [1 - 4 𝑓 (𝑥)] = - 2 𝑥 𝑓' (𝑥) = - 2 𝑥 1 - 4 𝑓 (𝑥) 𝑓' (𝑥) = 2 𝑥 4 𝑓 (𝑥) - 1

Insurance Process Step 4. Especially since the question asked us to find $𝑦',$ we'll substitute back $𝑓 (𝑥) = 𝑦$ and $𝑓' (𝑥) = 𝑦' .$

𝑦' = 2 𝑥 4 𝑦 - 1 ✓

If as you're practicing you find yourself ever missing the Chain Rule term $𝑑 𝑦 𝑑 𝑥$ or $𝑦',$ we strongly suggest using the $𝑦 = 𝑓 (𝑥)$ trick illustrated in Example 2. This simple move seems to help cue students to correctly apply the Chain Rule in high-stakes, exam situations.

Let's work through a Scaffolded Problem, where you can check your work at each key step. The question may appear intimidating at first, but the power of Implicit Differentiation is that it makes taking derivatives of even super-complicated-looking equations straightforward. And once that part is done, you're left with a standard "write the equation of a tangent line" problem, as you'll see.

Scaffolded Problem #1: Fifth degree polynomial

Graph of the fifth degree polynomial described by the given equation

Find the equation of the line tangent to the curve $𝑦 5 + 𝑥 𝑦 2 + 𝑥 = 𝑥 𝑦 4 + 𝑥 2 + 4 𝑦$ at the point $(- 3, - 2)$ .

Solution.

Step 1: Use implicit differentiation to find $𝑦' .$

(Note that in our solution, we will use the substitution $𝑦 = 𝑓 (𝑥)$ to make sure we don't forget to apply the Chain Rule.)

Step 2: Compute the slope of the tangent line at the point point $(- 3, - 2) .$

Step 3: Write the equation of the tangent line in point-slope form, $𝑦 - 𝑦 1 = 𝑚 (𝑥 - 𝑥 1) .$

Practice Problems

Time to practice! These problems can be kinda fun once you get the hang of them, and you'll probably find that after a few you have the routine down rather solidly. (But more practice never hurts!)

Practice Problem #1

Use implicit differentiation to find

𝑑 𝑦 𝑑 𝑥

given

𝑦 3 + 2 𝑥 2 + 𝑦 = 3 𝑥 4

(A) 4 𝑥 𝑦 (3 𝑥 2 - 1 1 + 3 𝑦) (B) (4 𝑥) 3 𝑥 2 - 1 1 + 3 𝑦 2 (C) 4 𝑥 3 𝑦 2 (3 𝑥 2 - 1)

(D) 4 𝑥 𝑦 (3 𝑥 2 - 1 𝑦 2 + 1) (E) N o n e o f t h e s e

ABCDE

Practice Problem #2

Use implicit differentiation to find

𝑦'

given

𝑒 𝑥 + c o s 𝑦 = 6 𝑦

(A) 𝑒 𝑥 6 + s i n 𝑦 (B) s i n - 1 (𝑒 𝑥 - 6) (C) c o s - 1 (6 - 𝑒 𝑥)

(D) 6 - 𝑒 𝑥 c o s 𝑦 (E) N o n e o f t h e s e

ABCDE

Practice Problem #3

Use implicit differentiation to find

𝑦'

given

𝑥 2 𝑦 2 = s i n 𝑦

(A) 𝑦' = 𝑥 c o s 𝑦 + 2 𝑥 2 𝑦 (B) 𝑦' = 2 𝑥 2 𝑦 s i n 𝑦 + 𝑦 2 c o s 𝑦 (C) 𝑦' = 𝑥 c o s 𝑦 + 2 s i n 𝑦

(D) 𝑦' = 𝑥 c o s 𝑦 + 𝑥 2 𝑦 (E) N o n e o f t h e s e

ABCDE

Practice Problem #4

Given

𝑥 4 [𝑓 (𝑥)] 2 = 4 𝑥 5 + s i n [𝑓 (𝑥)]

, find an expression for

𝑓' (𝑥)

(A) 𝑓' (𝑥) = 𝑥 3 [(𝑓 (𝑥)) 2 + 5 𝑥] c o s [𝑓 (𝑥)] - 2 𝑓 (𝑥) 𝑥 4 (B) 𝑓' (𝑥) = 4 𝑥 [(𝑓 (𝑥)) 2 - 5 𝑥] c o s [𝑓 (𝑥)] - 2 𝑓 (𝑥) 𝑥 4

(C) 𝑓' (𝑥) = 4 𝑥 3 [(𝑓 (𝑥)) 2 - 5 𝑥] c o s [𝑓 (𝑥)] - 2 𝑥 4 𝑓 (𝑥) (D) 𝑓' (𝑥) = 4 𝑥 3 [(𝑓 (𝑥)) 2 - 𝑥] c o s [𝑓 (𝑥)] - 𝑓 (𝑥) 𝑥 4 (E) N o n e o f t h e s e

ABCDE

Practice Problem #5

Given

𝑥 2 𝑔 (𝑥) + s i n 𝑥 = [𝑔 (𝑥)] 2

find

𝑔' (𝜋)

(A) 𝜋 2 - 1 (B) 1 - 1 𝜋 2 (C) 2 𝜋 - 1 𝜋 2 (D) 𝜋 2 - 1 2 𝜋 (E) N o n e o f t h e s e

ABCDE

Practice Problem #6

Consider an ellipse,

𝑥 2 𝑎 2 + 𝑦 2 𝑏 2 = 1 .

Find the slope of the curve

𝑦'

at the point

(𝑥, 𝑦) .

Bonus: Show that when

𝑎 = 𝑏,

the curve's slope reduces to

𝑦' = - 𝑥 𝑦

as we found above for the case of a circle.

(A) - 𝑎 2 𝑏 2 𝑥 𝑦 (B) - 𝑎 𝑏 𝑥 𝑦 (C) - 𝑏 𝑎 𝑥 𝑦 (D) - 𝑏 2 𝑎 2 𝑥 𝑦 (E) - 𝑎 3 𝑏 3 𝑥 𝑦

ABCDE

Practice Problem #7

Graph of x^2+xy+y^2 =7, which looks like a tilted ellipse centered at the origin.

Consider the relationship $𝑥 2 + 𝑥 𝑦 + 𝑦 2 = 7,$ which is shown in the graph. Find both values of $𝑦'$ for the curve at $𝑥 = 1$ .
$(A) - 45, - 15 (B) 53, 5 (C) 53, - 15 (D) - 45, 5 (E) N o n e o f t h e s e$

ABCDE

Practice Problem #8

Find

𝑦 ″

for the ellipse given by

𝑥 2 4 + 𝑦 2 9 = 1

(A) - 94 𝑥 𝑦 (B) - 94 [𝑦 - 𝑥 𝑦' 𝑦 2] (C) - 94

(D) - 94 [𝑦 2 + 94 𝑥 2 𝑦 3] (E) N o n e o f t h e s e

ABCDE

Practice Problem #9

𝑥 𝑦 + 𝑒 𝑦 = 𝑒

, find

𝑦 ″

𝑥 = 0

(A) 2 𝑒 (2 - 1 𝑒) (B) 1 𝑒 2 (C) 1 𝑒 (D) 2 𝑒 (1 - 1 𝑒) (E) 2 𝑒 2

ABCDE

Practice Problem #10

Find the equation of the line tangent to the graph of

𝑦 s i n 𝑥 = 𝑥 c o s 𝑦

at the point

(𝜋, 𝜋 2)

(A) 12 𝑥 (B) 𝜋 2 (𝑥 - 1) (C) 𝜋 (𝑥 2 - 1) (D) 𝜋 2 𝑥 (E) 12 (𝑥 + 1)

ABCDE

The Upshot

Using implicit differentiation to find the derivative of an implicitly defined function is straightforward: Step 1: Take the derivative $𝑑 𝑑 𝑥$ of both sides of the equation. The one thing you must be careful about: Remember the Chain Rule! Any term that includes a y with result in a Chain Rule term $𝑑 𝑦 𝑑 𝑥 .$ Step 2: Solve for $𝑑 𝑦 𝑑 𝑥$ .

Do you have an implicit differentiation question you're working on and could use some help with? Or any other comments or questions about what's on this screen? Please let us know on our Forum, where it's easy for you and for us to write the complicated math equations we're now working with.