Implicit Differentiation
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.8: “Implicit Differentiation” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Try This First
Without computing formally, estimate the slope of the circle $x^2 + y^2 = 25$ at the point $(3, 4)$.
Hint: the tangent to a circle is perpendicular to the radius. The radius to $(3, 4)$ has slope $4/3$.
Predicted slope: $\underline{\hspace{2cm}}$.
Now verify using implicit differentiation. The procedure should give you $-3/4$ at $(3, 4)$ -- which is $-1/(4/3)$, confirming that the tangent is perpendicular to the radius.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
The implicit differentiation procedure:
- Differentiate both sides of $F(x, y) = 0$ with respect to $x$.
- Every time a $y$-term is differentiated, multiply by $\dfrac{dy}{dx}$ (chain rule).
- Collect all $\dfrac{dy}{dx}$ terms on one side.
- Factor out $\dfrac{dy}{dx}$ and divide.
Key Concepts
1. The Full Procedure: The Circle
Example 1. Find $\dfrac{dy}{dx}$ for $x^2 + y^2 = 25$.
Step 1. Differentiate both sides with respect to $x$: \[ \frac{d}{dx}[x^2] + \frac{d}{dx}[y^2] = \frac{d}{dx}[25]. \]
Step 2. Apply known rules (chain rule on $y^2$): \[ 2x + 2y\frac{dy}{dx} = 0. \]
Step 3. Collect $\dfrac{dy}{dx}$ terms: \[ 2y\frac{dy}{dx} = -2x. \]
Step 4. Divide: \[ \frac{dy}{dx} = -\frac{x}{y}. \]
Check: This formula gives $-3/4$ at $(3, 4)$ and $-3/(-4) = 3/4$ at $(3, -4)$. Both make geometric sense: the upper arc slopes down from left to right at $(3,4)$, and the lower arc slopes up.
2. Mixed Terms: Product Rule + Chain Rule
Example 2. Find $\dfrac{dy}{dx}$ for $x^3 + y^3 = 6xy$ (folium of Descartes).
Differentiate: \[ 3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}. \]
Collect $dy/dx$ terms: \[ 3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2. \] \[ \frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2. \] \[ \frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}. \]
3. Implicit Differentiation with Trig and Exponentials
Example 3. Find $\dfrac{dy}{dx}$ for $\sin(xy) = y$.
Chain rule on left side (product inside): \[ \cos(xy) \cdot \frac{d}{dx}[xy] = \frac{dy}{dx}. \] \[ \cos(xy) \cdot \left(y + x\frac{dy}{dx}\right) = \frac{dy}{dx}. \] \[ y\cos(xy) + x\cos(xy)\frac{dy}{dx} = \frac{dy}{dx}. \] \[ y\cos(xy) = \frac{dy}{dx} - x\cos(xy)\frac{dy}{dx} = \frac{dy}{dx}(1 - x\cos(xy)). \] \[ \frac{dy}{dx} = \frac{y\cos(xy)}{1 - x\cos(xy)}. \]
4. Multiple Representations
Comparison of explicit and implicit approaches for $x^2 + y^2 = 25$ near $(3, 4)$:
Explicit: $y = \sqrt{25 - x^2}$. Then $\dfrac{dy}{dx} = \dfrac{-x}{\sqrt{25-x^2}}$. At $(3,4)$: $\dfrac{-3}{\sqrt{16}} = -\dfrac{3}{4}$.
Implicit: $\dfrac{dy}{dx} = -\dfrac{x}{y}$. At $(3,4)$: $-\dfrac{3}{4}$.
Same answer, but the implicit formula $-x/y$ works at every point on the circle simultaneously. The explicit formula requires choosing a branch first.
Graphical meaning: At $(3, 4)$, slope $= -3/4$. At $(0, 5)$ (top of circle), $dy/dx = -0/5 = 0$ (horizontal tangent). At $(5, 0)$ (rightmost point), $dy/dx = -5/0$ (undefined -- vertical tangent). The formula $-x/y$ encodes all of this.
5. Ask Why: Why Does Differentiating Both Sides Work?
If $F(x, y(x)) = c$ for all $x$ near $a$ (the equation holds along the curve), then differentiating both sides with respect to $x$ is valid: the left side is a function of $x$ (via the implicit function $y(x)$), and both sides equal $c$. Differentiating $c$ gives 0.
The chain rule on the left applies because $y$ is a function of $x$ -- even though we have not solved for it explicitly. As long as the curve is smooth near the point, $y(x)$ is differentiable and the chain rule applies.
Named Misconception: Forgetting to Collect $dy/dx$
After differentiating, students often solve for $dy/dx$ immediately from a single term instead of collecting all $dy/dx$ terms first. If $dy/dx$ appears in two places (as in Example 2), you must move all such terms to one side and factor before dividing.
Wrong procedure for $3y^2 \frac{dy}{dx} = 6y - 3x^2 + 6x\frac{dy}{dx}$:
“$dy/dx = (6y - 3x^2)/3y^2$” -- this ignores the $6x\,dy/dx$ term on the right.
Correct: move all $dy/dx$ terms to the left side first.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Missing $dy/dx$ on $y$-terms | $\frac{d}{dx}[y^3] = 3y^2$ | Must be $3y^2 \frac{dy}{dx}$ |
| Incorrectly solving for $dy/dx$ | Dividing by only one factor | Collect ALL terms with $dy/dx$ first, then factor and divide |
| Evaluating too early | Substituting specific $(x_0, y_0)$ before solving for $dy/dx$ | Find the general formula for $dy/dx$ first, then evaluate |
Common Misconceptions
differentiating a term containing $y$ with respect to $x$ produces the same result as differentiating a term containing $x$.
This is the composition-is-not-chaining error. Consider $\dfrac{d}{dx}[y^2]$. Because $y$ is a function of $x$, the chain rule requires multiplying the outer derivative by $\dfrac{dy}{dx}$: the correct result is $2y\dfrac{dy}{dx}$, not $2y$. Omitting this factor corrupts every subsequent step, because $\dfrac{dy}{dx}$ terms that should appear on one side of the equation are silently erased.
Leveled Practice
Level 1 -- Differentiating Both Sides
Problem 1. Find $\dfrac{dy}{dx}$ for $x^3 + 3y^2 = 10$.
Show answer
$3x^2 + 6y\dfrac{dy}{dx} = 0$.
$\dfrac{dy}{dx} = -\dfrac{x^2}{2y}$.
Problem 2. Find $\dfrac{dy}{dx}$ for $e^y = x + y$.
Show answer
$e^y \dfrac{dy}{dx} = 1 + \dfrac{dy}{dx}$.
$(e^y - 1)\dfrac{dy}{dx} = 1$.
$\dfrac{dy}{dx} = \dfrac{1}{e^y - 1}$.
Level 2 -- Mixed Terms
Problem 3. Find $\dfrac{dy}{dx}$ for $x^2 y + y^3 = 8$.
Show answer
Differentiate: $2xy + x^2\dfrac{dy}{dx} + 3y^2\dfrac{dy}{dx} = 0$.
$(x^2 + 3y^2)\dfrac{dy}{dx} = -2xy$.
$\dfrac{dy}{dx} = \dfrac{-2xy}{x^2 + 3y^2}$.
Problem 4. Find $\dfrac{dy}{dx}$ for $\cos(x + y) = y\sin x$.
Show answer
$-\sin(x+y)\left(1 + \dfrac{dy}{dx}\right) = \sin x\dfrac{dy}{dx} + y\cos x$.
$-\sin(x+y) - \sin(x+y)\dfrac{dy}{dx} = \sin x\dfrac{dy}{dx} + y\cos x$.
$-\sin(x+y) - y\cos x = (\sin x + \sin(x+y))\dfrac{dy}{dx}$.
$\dfrac{dy}{dx} = \dfrac{-\sin(x+y) - y\cos x}{\sin x + \sin(x+y)}$.
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 5 (Extension).
(a) (Floor) Verify that $(1, 1)$ lies on $x^3 + y^3 = 2$ and find $dy/dx$ at this point.
(b) (Mid) Use implicit differentiation on $x^n + y^n = c$ (where $n$ is a positive integer) to find a general formula for $dy/dx$ and describe the pattern.
(c) (Ceiling) Find $dy/dx$ for $y = x^x$ by first taking $\ln$ of both sides to get $\ln y = x\ln x$, then differentiating implicitly. What does this tell you about “logarithmic differentiation”?
Show answer
(a) $1^3 + 1^3 = 2$. Yes. Implicit: $3x^2 + 3y^2\dfrac{dy}{dx} = 0$, so $dy/dx = -x^2/y^2$. At $(1,1)$: $-1$.
(b) $nx^{n-1} + ny^{n-1}\dfrac{dy}{dx} = 0$, so $\dfrac{dy}{dx} = -\dfrac{x^{n-1}}{y^{n-1}}$.
For $n=2$ (circle): $-x/y$. For $n=3$: $-x^2/y^2$. The pattern is $-(x/y)^{n-1}$.
(c) $\ln y = x\ln x$. Differentiate: $\dfrac{1}{y}\dfrac{dy}{dx} = \ln x + x \cdot \dfrac{1}{x} = \ln x + 1$. So $\dfrac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$.
Logarithmic differentiation: take $\ln$ of both sides before differentiating. Simplifies products, quotients, and variable-exponent functions. Implicitly uses the chain rule on $\ln y$.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Implicit differentiation is the chain rule in disguise. You differentiate an equation in $x$ and $y$ with respect to $x$, treating $y$ as an unknown function $y(x)$. Every $y$-term generates a $dy/dx$ factor -- this is the chain rule at work. Collecting these factors and solving for $dy/dx$ extracts the slope formula without ever needing to solve for $y$ explicitly.
The key discipline: write $dy/dx$ next to every $y$-derivative, without exception. Omitting even one $dy/dx$ corrupts the entire calculation.
Connections
Within Calculus I (MATH161)
- Tangent lines: Once $dy/dx$ is found implicitly, evaluating at a specific point gives the slope of the tangent line to the curve. See the next lesson.
- Related rates: Related-rate problems differentiate implicit equations with respect to time $t$; the same chain-rule mechanics apply.
- Logarithmic differentiation: Implicitly differentiating $\ln y = \ln f(x)$ gives $dy/dx = f(x) \cdot (\ln f)'$, a technique for differentiating complex products and powers.
Back to Calculus I Skills | Previous: Implicitly Defined Functions | Next: Tangent Lines to Implicit Curves