Implicit Differentiation

Basic steps:

1. Differentiate both sides of the equation with respect to $x$ simultaneously
2. Treat $y$ as a function of $x$
3. Solve for $y'$

Principle (derivation):

Implicit differentiation is based on the chain rule. When the equation $F(x, y) = 0$ determines $y$ as a function of $x$, we have:

$\frac{d}{dx}F(x, y) = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \cdot \frac{dy}{dx} = 0$

Hence:

$\frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$

Geometric meaning:

The geometric meaning of implicit differentiation is that it gives the slope of the tangent line to a curve at a point. For example, for the circle $x^2 + y^2 = 1$, the slope of the tangent line at the point $(a, b)$ is $-\frac{a}{b}$.

Basic examples:

Example 1: the equation of a circle Find the derivative of the implicit function determined by $x^2 + y^2 = 1$ Solution:

• Differentiate both sides with respect to $x$: $2x + 2yy' = 0$
• Solve for $y'$: $y' = -\frac{x}{y}$

Example 2: a cubic curve Find the derivative of the implicit function determined by $x^3 + y^3 = 3xy$ Solution:

• Differentiate both sides with respect to $x$: $3x^2 + 3y^2y' = 3y + 3xy'$
• Rearrange: $3y^2y' - 3xy' = 3y - 3x^2$
• Solve for $y'$: $y' = \frac{y - x^2}{y^2 - x}$

Complicated examples:

Example 3: an exponential function Find the derivative of the implicit function determined by $e^{xy} + x^2 + y^2 = 1$ Solution:

• Differentiate both sides with respect to $x$: $e^{xy}(y + xy') + 2x + 2yy' = 0$
• Rearrange: $e^{xy}y + xe^{xy}y' + 2x + 2yy' = 0$
• Solve for $y'$: $y' = -\frac{e^{xy}y + 2x}{xe^{xy} + 2y}$

Example 4: a logarithmic function Find the derivative of the implicit function determined by $\ln(x^2 + y^2) = xy$ Solution:

• Differentiate both sides with respect to $x$: $\frac{2x + 2yy'}{x^2 + y^2} = y + xy'$
• Rearrange: $2x + 2yy' = (x^2 + y^2)(y + xy')$
• Solve for $y'$: $y' = \frac{(x^2 + y^2)y - 2x}{2y - (x^2 + y^2)x}$