How do I handle higher-order implicit derivatives like $d^2y/dx^2$?

Differentiate your expression for $dy/dx$ implicitly with respect to $x$ again. Then, substitute your first-derivative expression into the result to eliminate any $dy/dx$ terms from the final answer.

What if $dy/dx$ evaluates to division by zero at a point?

A zero in the denominator indicates that the tangent line to the curve is vertical at that specific point, meaning the derivative is undefined there.

Do I always have to algebraically isolate $dy/dx$ first?

No. If you are asked to find the derivative at a specific numerical point $(x_0, y_0)$, it is often faster and less error-prone to plug those values in immediately after differentiating, then solve for $dy/dx$.

How to differentiate implicitly

Implicit differentiation is a method used to find the derivative $dy/dx$ when a relationship between $x$ and $y$ is not easily solvable for $y$ in terms of $x$ .

It applies the chain rule by treating $y$ as an implicitly defined, differentiable function of $x$ .

The setup

Identify the equation relating $x$ and $y$ . Treat $x$ as the independent variable and $y$ as the dependent variable, $y(x)$ . Any operation performed on a term containing $y$ requires the chain rule when taking the derivative with respect to $x$ .

The steps

Differentiate both sides of the equation with respect to $x$ .
Apply the chain rule to all terms containing $y$ , which results in multiplying those terms' derivatives by $dy/dx$ .
Move all terms containing $dy/dx$ to one side of the equation and all other terms to the opposite side.
Factor out $dy/dx$ from the grouped terms.
Solve for $dy/dx$ by dividing both sides by the factored expression.

Checking the result

If evaluating at a specific point $(x_0, y_0)$ , plug the coordinates into the original equation to ensure the point actually lies on the curve. Alternatively, if the original equation can be solved explicitly for $y$ , differentiate the explicit function and compare it to your implicit result after substitution.

Common errors

Failing to apply the product rule to mixed terms (e.g., treating the derivative of $xy$ as just $y$ instead of $1 \\cdot y + x \\cdot \\frac{dy}{dx}$ ). Forgetting to multiply by $dy/dx$ when differentiating terms like $y^2$ .

Worked example

Find $dy/dx$ for the equation $x^2 + y^2 = 25$ .

$\\frac{d}{dx}(x^2 + y^2) = \\frac{d}{dx}(25)$ $2x + 2y \\frac{dy}{dx} = 0$ $2y \\frac{dy}{dx} = -2x$ $\\frac{dy}{dx} = -\\frac{2x}{2y}$ $\\frac{dy}{dx} = -\\frac{x}{y}$

FAQ

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Answer form

References: Stewart Calculus, Chapter 3 · OpenStax Calculus Volume 1, Section 3.8 · Khan Academy: Implicit differentiation