Implicit Differentiation Revision Notes for Leaving Cert Mathematics

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of the other. Instead of solving the equation for y in terms of $x$ first, you differentiate both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ .

Implicit Equations: Implicit differentiation is used when the equation is given in a form where $y$ is not isolated on one side (e.g., $x^2 + y^2 = 1, x^3 + y^3 = 3xy$ ).

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Steps for Implicit Differentiation

Differentiate both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ . When differentiating a term involving $y$ , apply the chain rule ( $\frac{d}{dx} y = \frac{dy}{dx})$ .
Collect terms involving $\frac{dy}{dx}$ on one side of the equation.
Solve for $\frac{dy}{dx}$ to find the derivative.

Example

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Differentiate $x^2 + y^2 = 1$

Differentiate both sides with respect to $x$ :

\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(1)

2x + 2y \frac{dy}{dx} = 0

Solve for $\frac{dy}{dx}:$

2y \frac{dy}{dx} = -2x

\frac{dy}{dx} = \frac{-x}{y}

The derivative of $y$ with respect to $x$ for the circle is $\frac{dy}{dx} = \frac{-x}{y}.$

Example

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Differentiate $x^3 + y^3 = 6xy$

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( x \frac{dy}{dx} + y \right)

Distribute and collect terms involving $\frac{dy}{dx}$

3y^2 \frac{dy}{dx} - 6x \frac{dy}{dx} = 6y - 3x^2

\frac{dy}{dx} (3y^2 - 6x) = 6y - 3x^2

Solve for $\frac{dy}{dx}:$

\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}

The derivative of $y$ with respect to $x$ for the given equation is $\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}.$

Implicit Differentiation (Leaving Cert Mathematics): Revision Notes