Implicit Differentiation for MATH 139
Exam Relevance for MATH 139
Differentiate both sides with respect to x, use chain rule on y terms, then solve for dy/dx.
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What is Implicit Differentiation?
Implicit differentiation is a technique for finding $\frac{dy}{dx}$ when $y$ is not isolated on one side of the equation.
Explicit form: $y = x^2 + 3x$ (y is isolated — easy to differentiate)
Implicit form: $x^2 + y^2 = 25$ (y is mixed with x — can't easily solve for y)
When you can't (or don't want to) solve for $y$ explicitly, use implicit differentiation!
The Key Idea
When differentiating with respect to $x$:
- Treat $x$ terms normally
- Treat $y$ as a function of $x$, so use the chain rule whenever you differentiate $y$
$$\frac{d}{dx}[y^2] = 2y \cdot \frac{dy}{dx}$$
The $\frac{dy}{dx}$ appears because $y$ depends on $x$ (chain rule!).
The Process
- Differentiate both sides with respect to $x$
- Apply chain rule to any term containing $y$ (multiply by $\frac{dy}{dx}$)
- Collect all $\frac{dy}{dx}$ terms on one side
- Factor out $\frac{dy}{dx}$
- Solve for $\frac{dy}{dx}$
Problem: Find $\frac{dy}{dx}$ 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 the derivatives (chain rule on $y^2$): $$2x + 2y \cdot \frac{dy}{dx} = 0$$
Step 3: Solve for $\frac{dy}{dx}$: $$2y \cdot \frac{dy}{dx} = -2x$$ $$\frac{dy}{dx} = \frac{-2x}{2y} = \boxed{-\frac{x}{y}}$$
Problem: Find $\frac{dy}{dx}$ for $xy = 1$
Step 1: Use the product rule on $xy$: $$\frac{d}{dx}[xy] = \frac{d}{dx}[1]$$ $$\frac{d}{dx}[x] \cdot y + x \cdot \frac{d}{dx}[y] = 0$$
Step 2: Apply derivatives: $$1 \cdot y + x \cdot \frac{dy}{dx} = 0$$ $$y + x\frac{dy}{dx} = 0$$
Step 3: Solve for $\frac{dy}{dx}$: $$x\frac{dy}{dx} = -y$$ $$\frac{dy}{dx} = \boxed{-\frac{y}{x}}$$
Problem: Find $\frac{dy}{dx}$ for $x^3 + y^3 = 6xy$
Step 1: Differentiate both sides: $$\frac{d}{dx}[x^3] + \frac{d}{dx}[y^3] = \frac{d}{dx}[6xy]$$
Step 2: Apply derivatives (product rule on right side): $$3x^2 + 3y^2 \cdot \frac{dy}{dx} = 6\left(1 \cdot y + x \cdot \frac{dy}{dx}\right)$$ $$3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$$
Step 3: Collect $\frac{dy}{dx}$ terms on one side: $$3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2$$
Step 4: Factor out $\frac{dy}{dx}$: $$\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$$
Step 5: Solve: $$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{3(2y - x^2)}{3(y^2 - 2x)} = \boxed{\frac{2y - x^2}{y^2 - 2x}}$$
Problem: Find $\frac{dy}{dx}$ for $\sin(xy) = x$
Step 1: Differentiate both sides (chain rule on left): $$\cos(xy) \cdot \frac{d}{dx}[xy] = 1$$
Step 2: Apply product rule to $xy$: $$\cos(xy) \cdot \left(y + x\frac{dy}{dx}\right) = 1$$
Step 3: Distribute: $$y\cos(xy) + x\cos(xy)\frac{dy}{dx} = 1$$
Step 4: Solve for $\frac{dy}{dx}$: $$x\cos(xy)\frac{dy}{dx} = 1 - y\cos(xy)$$ $$\frac{dy}{dx} = \boxed{\frac{1 - y\cos(xy)}{x\cos(xy)}}$$
Problem: Find the slope of the tangent line to $x^2 + y^2 = 25$ at the point $(3, 4)$.
Step 1: We already found $\frac{dy}{dx} = -\frac{x}{y}$ in Example 1.
Step 2: Substitute the point $(3, 4)$: $$\frac{dy}{dx}\bigg|_{(3,4)} = -\frac{3}{4}$$
$$\boxed{\text{slope} = -\frac{3}{4}}$$
Problem: Find $\frac{d^2y}{dx^2}$ for $x^2 + y^2 = 25$
Step 1: We know $\frac{dy}{dx} = -\frac{x}{y}$
Step 2: Differentiate again using the quotient rule: $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[-\frac{x}{y}\right] = -\frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2}$$
Step 3: Substitute $\frac{dy}{dx} = -\frac{x}{y}$: $$= -\frac{y - x \cdot \left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2}$$
Step 4: Simplify (multiply numerator and denominator by $y$): $$= -\frac{y^2 + x^2}{y^3}$$
Step 5: Since $x^2 + y^2 = 25$: $$\frac{d^2y}{dx^2} = \boxed{-\frac{25}{y^3}}$$
When to Use Implicit Differentiation
- When the equation can't be easily solved for $y$ (e.g., $x^3 + y^3 = 6xy$)
- When solving for $y$ would give multiple branches (e.g., $y = \pm\sqrt{25-x^2}$)
- When you want the derivative in terms of both $x$ and $y$
- Always works, even when explicit differentiation is possible!
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting the chain rule on $y$ terms
Wrong: $\frac{d}{dx}[y^2] = 2y$
Why it's wrong: Since $y$ is a function of $x$, you must apply the chain rule.
Correct: $\frac{d}{dx}[y^2] = 2y \cdot \frac{dy}{dx}$
❌ Mistake: Forgetting the product rule when $x$ and $y$ are multiplied
Wrong: $\frac{d}{dx}[xy] = \frac{dy}{dx}$
Why it's wrong: $xy$ is a product of two functions of $x$. Use the product rule!
Correct: $\frac{d}{dx}[xy] = 1 \cdot y + x \cdot \frac{dy}{dx} = y + x\frac{dy}{dx}$
❌ Mistake: Not collecting all $\frac{dy}{dx}$ terms before solving
Wrong: Trying to solve for $\frac{dy}{dx}$ when it appears on both sides.
Why it's wrong: You'll get confused and make algebra errors.
Correct: First move ALL terms with $\frac{dy}{dx}$ to one side, then factor it out.
Quick Reference
| Expression | Derivative with respect to $x$ |
|---|---|
| $y$ | $\frac{dy}{dx}$ |
| $y^2$ | $2y\frac{dy}{dx}$ |
| $y^n$ | $ny^{n-1}\frac{dy}{dx}$ |
| $xy$ | $y + x\frac{dy}{dx}$ |
| $\sin y$ | $\cos y \cdot \frac{dy}{dx}$ |
| $e^y$ | $e^y \cdot \frac{dy}{dx}$ |
| $\ln y$ | $\frac{1}{y} \cdot \frac{dy}{dx}$ |
Chain Rule for y Terms
When differentiating y^n with respect to x, apply the power rule AND multiply by dy/dx (chain rule).
Variables:
- $y$:
- a function of x (dependent variable)
- $n$:
- the exponent
- $\frac{dy}{dx}$:
- the derivative we're solving for
Product Rule for xy
When x and y are multiplied, use the product rule. The derivative of x is 1, and the derivative of y is dy/dx.
Variables:
- $x$:
- independent variable
- $y$:
- dependent variable (function of x)
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