So far, we've differentiated functions written explicitly as $y = f(x)$. But many curves — like circles $x^2 + y^2 = 25$ — are not functions. Yet we can still find $\frac{dy}{dx}$ using implicit differentiation. We treat $y$ as a function of $x$ and differentiate both sides of the equation, then solve for $dy/dx$.
🎯 In this section you will learn
- What implicit differentiation is and when to use it
- How to differentiate $y$ as a function of $x$ (Chain Rule!)
- Finding $\frac{dy}{dx}$ for circles, ellipses, and other relations
- Finding slopes of tangent lines to implicit curves
📌 The Key Idea
When we differentiate an equation like $x^2 + y^2 = 25$, we treat $y$ as $y(x)$ — a function of $x$. So:
$$ \frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx} \quad \text{(Chain Rule!)} $$
💡 Important
Whenever you differentiate a term involving $y$, you multiply by $\frac{dy}{dx}$ (because $y$ is a function of $x$).
- $\frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx}$
- $\frac{d}{dx}(y^5) = 5y^4 \cdot \frac{dy}{dx}$
- $\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}$
- $\frac{d}{dx}(e^y) = e^y \cdot \frac{dy}{dx}$
- $\frac{d}{dx}(xy) = x \cdot \frac{dy}{dx} + y \cdot 1$ (Product Rule!)
📌 Steps for Implicit Differentiation
①
Differentiate both sides
Differentiate every term with respect to $x$. Remember: derivative of $y$ is $\frac{dy}{dx}$.
②
Collect $\frac{dy}{dx}$ terms
Move all terms containing $\frac{dy}{dx}$ to one side.
③
Factor out $\frac{dy}{dx}$
Factor $\frac{dy}{dx}$ from the collected terms.
④
Solve for $\frac{dy}{dx}$
Divide both sides by the coefficient of $\frac{dy}{dx}$.
Find $\frac{dy}{dx}$ for the circle $x^2 + y^2 = 25$.
①
Differentiate both sides
$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$
②
Apply the rules
$2x + 2y \cdot \frac{dy}{dx} = 0$
③
Solve for $\frac{dy}{dx}$
$2y \cdot \frac{dy}{dx} = -2x \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}$
Find the slope of the tangent line to $x^2 + y^2 = 25$ at the point $(3, -4)$.
①
Use $\frac{dy}{dx} = -\frac{x}{y}$
From Example 2, $\frac{dy}{dx} = -\frac{x}{y}$
②
Plug in $(3, -4)$
$\frac{dy}{dx} = -\frac{3}{-4} = \frac{3}{4}$
③
Interpret
The slope of the tangent line at $(3, -4)$ is $\frac{3}{4}$.
Find $\frac{dy}{dx}$ for $x^2 + xy + y^2 = 7$.
①
Differentiate
$\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = 0$
②
Apply rules
$2x + \left(x \cdot \frac{dy}{dx} + y \cdot 1\right) + 2y \cdot \frac{dy}{dx} = 0$
③
Collect $\frac{dy}{dx}$ terms
$2x + y + x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0$
④
Factor and solve
$(x + 2y)\frac{dy}{dx} = -2x - y \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{2x + y}{x + 2y}$
Find $\frac{dy}{dx}$ for $\sin x + \cos y = 1$.
①
Differentiate
$\frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos y) = 0$
②
Apply rules
$\cos x - \sin y \cdot \frac{dy}{dx} = 0$
③
Solve
$\frac{dy}{dx} = \frac{\cos x}{\sin y}$
Find $\frac{dy}{dx}$ for $e^x + e^y = xy$.
①
Differentiate
$e^x + e^y \cdot \frac{dy}{dx} = x \cdot \frac{dy}{dx} + y \cdot 1$
②
Collect $\frac{dy}{dx}$ terms
$e^y \frac{dy}{dx} - x \frac{dy}{dx} = y - e^x$
③
Factor and solve
$(e^y - x) \frac{dy}{dx} = y - e^x \quad \Rightarrow \quad \frac{dy}{dx} = \frac{y - e^x}{e^y - x}$
⚠️ Common Mistakes
- Forgetting the $\frac{dy}{dx}$ factor: $\frac{d}{dx}(y^2) = 2y$, not $2y \frac{dy}{dx}$ — wait, that's the mistake! It SHOULD be $2y \frac{dy}{dx}$.
- Applying the Product Rule to $xy$ incorrectly: Remember: $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y$.
- Not simplifying after solving: Always simplify your final expression for $\frac{dy}{dx}$ if possible.
🔍 Key Takeaways
- Implicit differentiation is used when $y$ is not isolated (e.g., circles, ellipses).
- Every time you differentiate $y$, multiply by $\frac{dy}{dx}$ (Chain Rule).
- Use the Product Rule for terms like $xy$.
- After differentiating, collect $\frac{dy}{dx}$ terms and solve.
- You can find slopes at specific points without solving for $y$ explicitly.
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