4.5 Implicit Differentiation

4.4 Implicit Differentiation

Quick Answer: When an equation defines a relation between $x$ and $y$ implicitly ($F(x, y) = 0$), differentiate both sides of the equation with respect to $x$, treating $y$ as an unknown differentiable function of $x$. Whenever differentiating a term containing $y$, apply the Chain Rule and attach a factor of $\frac{dy}{dx}$ (or $y'$). Then collect all $\frac{dy}{dx}$ terms on one side, factor out $\frac{dy}{dx}$, and solve algebraically.

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1. Explicit Functions vs. Implicit Relations

• Explicit Function: The dependent variable is completely isolated on one side: $y = f(x)$, such as $y = 3x^2 - 5x + 2$.
• Implicit Relation: The variables $x$ and $y$ are intertwined in an equation: $F(x, y) = C$, such as the circle $x^2 + y^2 = 25$, the ellipse $4x^2 + 9y^2 = 36$, or the Folium of Descartes $x^3 + y^3 = 6xy$.

In many implicit equations, solving for $y$ explicitly in terms of $x$ is either algebraically impossible or introduces complicated piecewise radical expressions. Implicit differentiation provides a direct pathway to finding $\frac{dy}{dx}$ without solving for $y$.

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2. The Underlying Chain Rule Mechanism

Because $y$ represents an implicit function of $x$ ($y = y(x)$), any operation applied to $y$ must be differentiated using the Chain Rule with respect to the independent variable $x$.

                     Differentiating Terms with Respect to x

      Term with x alone:       d/dx [ x³ ]       = 3x²
      Term with y alone:       d/dx [ y³ ]       = 3y² · (dy/dx)
      Mixed Product Term:      d/dx [ x² y³ ]    = 2x y³ + x² (3y² · dy/dx)  [Product Rule]
      Composite Trig Term:     d/dx [ sin(xy) ]  = cos(xy) · (y + x · dy/dx)  [Chain + Product]

Essential Derivative Building Blocks with Respect to $x$

• Power of $y$: $\frac{d}{dx}[y^n] = n y^{n-1} \frac{dy}{dx}$
• Product $xy$: $\frac{d}{dx}[xy] = (1)y + x\left(\frac{dy}{dx}\right) = y + x y'$
• Product $x^2 y$: $\frac{d}{dx}[x^2 y] = 2xy + x^2 y'$
• Trigonometric term $\sin(y)$: $\frac{d}{dx}[\sin y] = \cos y \cdot y'$
• Exponential term $e^{xy}$: $\frac{d}{dx}[e^{xy}] = e^{xy}(y + x y')$

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3. The 4-Step Systematic Algorithm for Isolating $\frac{dy}{dx}$

1. Step 1 (Differentiate): Take $\frac{d}{dx}$ of every term on both sides of the equation. Remember that the derivative of a constant is $0$.
2. Step 2 (Expand): Multiply out all parentheses so that every term is an isolated additive term.
3. Step 3 (Collect): Move all terms containing $\frac{dy}{dx}$ to the left-hand side of the equation, and move all terms without $\frac{dy}{dx}$ to the right-hand side.
4. Step 4 (Factor & Solve): Factor out $\frac{dy}{dx}$ on the left and divide by its multiplying coefficient.

Full Worked Example: Folium of Descartes

Find $\frac{dy}{dx}$ for $x^3 + y^3 = 6xy$.

1. Differentiate both sides with respect to $x$: $$\frac{d}{dx}[x^3] + \frac{d}{dx}[y^3] = \frac{d}{dx}[6xy]$$ $$3x^2 + 3y^2 \frac{dy}{dx} = 6\left( y + x \frac{dy}{dx} \right)$$
2. Expand right side: $$3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}$$
3. Collect $\frac{dy}{dx}$ terms on the left: $$3y^2 \frac{dy}{dx} - 6x \frac{dy}{dx} = 6y - 3x^2$$
4. Factor out $\frac{dy}{dx}$ and divide: $$\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$$ $$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$$

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4. Tangent and Normal Lines to Implicit Curves

To find the equation of a line tangent or normal to an implicit curve at a given point $(x_0, y_0)$:

CLEP Time-Saving Strategy: Do not waste time solving for the algebraic expression of $\frac{dy}{dx}$ first! Differentiate the equation implicitly, then immediately plug in the numerical coordinates $x = x_0$ and $y = y_0$. This converts the equation into a simple linear algebraic equation in terms of $y'$.

Worked Example: Tangent and Normal Lines to $x^2 - xy + y^2 = 7$ at $(1, 3)$

1. Verify the point lies on the curve: $1^2 - (1)(3) + 3^2 = 1 - 3 + 9 = 7$ (valid).
2. Differentiate implicitly: $$2x - (y + x y') + 2y y' = 0$$ $$2x - y - x y' + 2y y' = 0$$
3. Substitute $x = 1$ and $y = 3$ immediately: $$2(1) - 3 - (1)y' + 2(3)y' = 0$$ $$-1 - y' + 6y' = 0 \implies 5y' = 1 \implies m_{\text{tan}} = y' = \frac{1}{5}$$
4. Write Tangent Line Equation: $$y - 3 = \frac{1}{5}(x - 1) \implies y = \frac{1}{5}x + \frac{14}{5}$$
5. Write Normal Line Equation: The normal line is perpendicular to the tangent line: $m_{\text{norm}} = -\frac{1}{m_{\text{tan}}} = -5$. $$y - 3 = -5(x - 1) \implies y = -5x + 8$$

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5. Finding Horizontal and Vertical Tangents

For an implicit derivative expressed as a rational fraction $\frac{dy}{dx} = \frac{N(x, y)}{D(x, y)}$:

Worked Example: Ellipse $x^2 + 4y^2 = 16$

1. Differentiate implicitly: $2x + 8y y' = 0 \implies y' = -\frac{2x}{8y} = -\frac{x}{4y}$.
2. Horizontal Tangents: Set numerator $-x = 0 \implies x = 0$.
   • Substitute $x = 0$ into ellipse: $0^2 + 4y^2 = 16 \implies y^2 = 4 \implies y = \pm 2$.
   • Points with horizontal tangents: $(0, 2)$ and $(0, -2)$.
3. Vertical Tangents: Set denominator $4y = 0 \implies y = 0$.
   • Substitute $y = 0$ into ellipse: $x^2 + 4(0)^2 = 16 \implies x^2 = 16 \implies x = \pm 4$.
   • Points with vertical tangents: $(4, 0)$ and $(-4, 0)$.

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6. Higher-Order Implicit Derivatives ($\frac{d^2 y}{dx^2}$)

Finding the second derivative of an implicit relation requires two consecutive differentiations followed by back-substitution.

Classic Derivation: Second Derivative of Circle $x^2 + y^2 = r^2$

1. First derivative: $$2x + 2y y' = 0 \implies y' = -\frac{x}{y}$$
2. Differentiate $y'$ with respect to $x$ using the Quotient Rule: $$y'' = \frac{d}{dx}\left[ -\frac{x}{y} \right] = -\frac{(1)(y) - x(y')}{y^2} = -\frac{y - x y'}{y^2}$$
3. Back-substitute $y' = -\frac{x}{y}$ into the numerator: $$y'' = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2}$$
4. Clear the compound fraction by multiplying top and bottom by $y$: $$y'' = -\frac{y^2 + x^2}{y^3}$$
5. Back-substitute the original circle equation $x^2 + y^2 = r^2$: $$y'' = -\frac{r^2}{y^3}$$

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7. CLEP Traps and Exam Strategies

• Trap 1: The "Missing Product Rule" on $xy$. Differentiating $xy$ as simply $y'$ or $1 \cdot y'$ is the most common error in calculus. Always write $\frac{d}{dx}[xy] = y + x y'$.
• Trap 2: Forgetting to differentiate constants on the RHS. Writing $\frac{d}{dx}[x^2 + y^2 = 25] \implies 2x + 2y y' = 25$ ruins the entire solution. The derivative of any constant is $0$.
• Trap 3: Forgetting to verify points on the original curve. When solving for horizontal or vertical tangents, you must plug the resulting condition back into the original relation to find valid coordinate points.