Implicit Differentiation, Explained
x and y share an equation that is awkward to solve for y.
Differentiate both sides, attach y′ to y-terms, isolate y′.
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
- Differentiate both sides: apply \(\frac{d}{dx}\) to every term.
- Apply differentiation rules: apply the chain rule when differentiating y(x), together with the product or quotient rule for mixed terms.
- Isolate the \(\frac{dy}{dx}\) terms: use algebra to gather them on one side.
- 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: \(\frac{dy}{dx} = -\frac{x}{y}\)
Worked implicit derivatives: three different cases
Mixed powers and products: \(x^2+xy+y^2=7\)
Differentiate each term with respect to x. The product xy needs the product rule, and y² needs the chain rule:
\[2x+y+xy'+2yy'=0.\]
Collect and factor the derivative terms: \((x+2y)y'=-(2x+y)\). Therefore, where \(x+2y\ne0\),
\[y'=-\frac{2x+y}{x+2y}.\]
The point (1,2) lies on the curve because 1+2+4=7. Its slope is −4/5, so the tangent line is \(y-2=-\frac45(x-1)\). Substituting a point only after differentiation keeps the variables available for the rules.
A trigonometric relation: \(\sin y=x\)
Because y depends on x, differentiating gives \(\cos y\,y'=1\). Thus \(y'=1/\cos y\) wherever \(\cos y\ne0\).
On the principal branch \(y=\arcsin x\), −1<x<1 and \(\cos y=\sqrt{1-x^2}\). This gives \(y'=1/\sqrt{1-x^2}\). Other branches can have a different sign, so do not replace cos y with a positive square root without choosing a branch.
At (0,0), the slope is 1 and the tangent is y=x. Compare this with the worked derivative of arcsin x.
A singular point: \(y^2=x^2\)
Differentiation gives \(2yy'=2x\), hence \(y'=x/y\) when y≠0. At the origin this ratio is 0/0, which does not determine a slope.
Factor the original equation: \((y-x)(y+x)=0\). The relation contains two branches, y=x and y=−x, with slopes 1 and −1. Each branch has its own tangent at the origin; the full relation has no single tangent slope there. A zero denominator alone does not prove a vertical tangent.
For a differentiable relation F(x,y)=0, the compact rule is \(y'=-F_x/F_y\) wherever \(F_y\ne0\) and a differentiable branch exists. Here \(F_x\) differentiates with respect to x while holding y fixed; \(F_y\) does the reverse. The step-by-step method above derives the same formula without requiring partial-derivative notation.
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.
Valid points and calculator scope
For x² + y² = 25, the formulas y′ = −x/y and y″ = −25/y³ apply on differentiable branches where y ≠ 0. At (3, 4), the tangent line is y − 4 = −3(x − 3)/4. Always check the proposed point satisfies the original equation before evaluating a slope.
At (5, 0) and (−5, 0), the circle has vertical tangents x = 5 and x = −5. The formula for dy/dx cannot supply a finite slope there. Division by a vanishing coefficient can also indicate a singular point on other curves; inspect the original relation.
Mixed terms need the full differentiation rules. For xy = 6, the product rule gives y + xy′ = 0, so y′ = −y/x. Simply attaching y′ to the entire product would miss the y term.
This page teaches implicit differentiation. The current calculator differentiates supported explicit expressions; it does not solve general equations involving x and y. For the upper circle branch, you can enter sqrt(25-x^2) on −5 < x < 5 and compare its explicit derivative with −x/y.
Check your understanding
- Find y′ if x²+y²=16.
Show answer
y′=−x/y where y≠0.
- Find y′ if xy=6.
Show answer
y′=−y/x.
- 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.