DerivCalc
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Implicit Differentiation, Explained

Find dy/dx when x and y share an equation, with a four-step method, tangent slopes, and higher derivatives.

Guide 4 · Techniques

Implicit Differentiation, Explained

Written by SilverCodeLabsReviewed for mathematical clarityRevised August 27, 2026
Recognize

x and y share an equation that is awkward to solve for y.

Apply

Differentiate both sides, attach y′ to y-terms, isolate y′.

Verify

Verify the point lies on the curve before finding its slope.

In calculus, we often work with functions like \(y = x^2\). But what about a circle, \(x^2 + y^2 = 25\)? Solving for \(y\) is messy. Implicit differentiation is the key that unlocks the calculus of curves — not just functions — allowing us to find \(\frac{dy}{dx}\) without ever isolating \(y\).

The Core Technique: A Step-by-Step Guide

The process hinges on one idea: treat \(y\) as a function of \(x\), written \(y(x)\). When we differentiate a term like \(y^2\) with respect to \(x\), we must use the chain rule: \(\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}\).

The four-step process

  1. Differentiate both sides: apply \(\frac{d}{dx}\) to every term.
  2. Apply differentiation rules: for every term involving \(y\), multiply by \(\frac{dy}{dx}\).
  3. Isolate the \(\frac{dy}{dx}\) terms: use algebra to gather them on one side.
  4. Solve for \(\frac{dy}{dx}\) by factoring and dividing.

Example: differentiating the circle \(x^2 + y^2 = 25\)

Differentiate: \(2x + 2y\frac{dy}{dx} = 0\)

Isolate: \(2y\frac{dy}{dx} = -2x\)

Solve: \(\frac{dy}{dx} = -\frac{x}{y}\)

Higher-Order Implicit Derivatives

We can find the second derivative \(\frac{d^2y}{dx^2}\) by differentiating the first derivative — with one crucial substitution step along the way.

Finding \(\frac{d^2y}{dx^2}\) for the circle

1. First derivative: \(\frac{dy}{dx} = -\frac{x}{y}\)

2. Differentiate again (quotient rule): \(\frac{d^2y}{dx^2} = -\frac{(1)y - x\left(\frac{dy}{dx}\right)}{y^2}\)

3. Substitute \(\frac{dy}{dx}\): \(\frac{d^2y}{dx^2} = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2}\)

4. Simplify using \(x^2+y^2=25\): \(\frac{d^2y}{dx^2} = -\frac{x^2+y^2}{y^3} = -\frac{25}{y^3}\)

Applications and Related Techniques

Implicit differentiation is crucial for:

  • Finding tangent and normal lines to complex curves.
  • Solving related rates problems, where variables change with respect to time.
  • Logarithmic differentiation, a technique for handling functions like \(y=x^x\) by first taking the natural log of both sides.

Implicit differentiation liberates us from the rigid constraint of explicit functions. By mastering this process — built on careful application of the chain rule — we gain a more powerful lens for analyzing the interconnected, dynamic curves of mathematics.

Check your understanding

  1. Find y′ if x²+y²=16.
    Show answer

    y′=−x/y where y≠0.

  2. Find y′ if xy=6.
    Show answer

    y′=−y/x.

  3. Find the slope on x²+y²=25 at (3,4).
    Show answer

    −3/4.

Reference standard: Rule statements and notation follow OpenStax Calculus Volume 1, Chapter 3. DerivCalc's explanations and examples are independently written.

Editorial note: Examples are checked symbolically and with numerical sampling. Read the methodology and limitations or report a correction.