Calculus I · Unit 2A · lesson

Implicit Differentiation

Concept

Learning objectives

Differentiate implicit relations and solve algebraically for $dy/dx$ .

Implicit, Inverse, and Logarithmic Differentiation

Differentiate an Equation Without Solving for $y$

Explanation

Before the formulas

In Implicit Differentiation, inverse and implicit ideas meet. Swapping input and output swaps horizontal and vertical change, so inverse slopes are reciprocals at corresponding points. Taking logarithms can also reveal hidden structure by turning products into sums and exponents into coefficients.

These methods are strategic transformations, not new definitions of derivative. State the domain assumptions, preserve the original relationship, and substitute back at the end. A clean solution explains why the transformation helps before carrying out the algebra.

**Visual study**
The circle x²+y²=25 contains upper and lower branches. Implicit differentiation gives one slope formula, dy/dx=-x/y, valid wherever y 0.

Explanation

Differentiate the relationship even when $y$ is not isolated

An equation such as $x^2+y^2=25$ describes a curve without giving one global formula for $y$ . Implicit differentiation treats $y$ as a function of $x$ and differentiates both sides of the relationship.

Whenever a derivative passes through an expression containing $y$ , the chain rule contributes $dy/dx$ . That factor records the fact that $y$ changes when $x$ changes. After differentiating, collect the $dy/dx$ terms and solve for the slope.

Not every curve is naturally written as $y=f(x)$ . Circles, ellipses, thermodynamic constraints, and many geometric relations describe $x$ and $y$ together. Implicit differentiation treats $y$ as a function of $x$ locally, even when solving explicitly would be awkward or would split the curve into branches.

Every time a derivative passes through an expression involving $y$ , the chain rule contributes a factor $dy/dx$ . That factor is the algebraic trace of the hidden dependence $y=y(x)$ .

An explicit equation gives $y$ directly as a function of $x$ , such as $y=x^2+1$ . An implicit equation relates $x$ and $y$ without isolating $y$ , such as

x^2+y^2=25.

Solving the circle for $y$ creates two branches. Implicit differentiation handles both at once.

When differentiating with respect to $x$ , remember that $y$ is itself a function of $x$ . Therefore

\frac{d}{dx}[y^n]=ny^{n-1}\frac{dy}{dx}.

The factor $dy/dx$ is the chain rule recording that $y$ changes when $x$ changes.

Guided walkthrough

Slope on a circle

Find $dy/dx$ for

x^2+y^2=25.

Answer reveal

Worked solution

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Worked example

An equation with products of $x$ and $y$

For

x^2y+y^3=6,

differentiate the product $x^2y$ with the product rule:

2xy+x^2y'+3y^2y'=0.

Group the derivative terms:

(x^2+3y^2)y'=-2xy.

Thus

\boxed{y'=-\frac{2xy}{x^2+3y^2}}.

Common mistake

The derivative of $y^2$ with respect to $x$ is not merely $2y$ . It is $2yy'$ . Omitting $y'$ pretends that $y$ is an independent constant while simultaneously trying to calculate how it changes.

Interactive checkimplicit-01

For $x^2+4y^2=20$ , find $dy/dx$ .

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Show hint

Differentiate y^2 with a factor y' and solve for y'.

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Modeling lab

Pressure and volume of a compressed gas

For a fixed amount of gas at constant temperature, suppose

PV=600,

with pressure in kilopascals and volume in liters. Treat $P$ as a function of $V$ . Differentiating gives

P+V\frac{dP}{dV}=0, \qquad \frac{dP}{dV}=-\frac{P}{V}.

At $V=20$ , $P=30$ , so $dP/dV=-1.5$ kPa/L. A small increase in volume near this state lowers pressure by about $1.5$ kPa per liter.

Optional advanced note

The Implicit Function Theorem hiding underneath

For an equation $F(x,y)=0$ , implicit differentiation formally gives

F_x+F_y\frac{dy}{dx}=0, \qquad \frac{dy}{dx}=-\frac{F_x}{F_y}.

A later theorem explains when this is legitimate: if $F$ is sufficiently smooth and $F_y\ne0$ at the point, then the equation really does define $y$ as a differentiable function of $x$ nearby. The denominator condition is the local solvability condition.

After the explanation

Use the section idea

Reading lens

Track which variable depends on which and use reciprocal or logarithmic structure only where its conditions hold.

Mental model

Implicit equations constrain variables together; inverse functions exchange inputs and outputs; logarithms turn products and powers into sums.

Decision

Choose implicit, inverse, or logarithmic differentiation from the equation's representation, not from surface complexity.

Common trap

Dropping a y-prime factor, using a reciprocal slope at the wrong point, or ignoring domain restrictions.

Check yourself

Can you identify the correspondence point and all hidden dependencies before differentiating?

Interactive checkimplicit-extra-01

For $x^2+y^2=9$ , find $dy/dx$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Differentiate $y^2$ as $2yy'$ .

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Source & rights

Original instruction with traceable references.

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Implicit, Inverse, and Logarithmic Differentiation

Differentiate an Equation Without Solving for yyy

Original instruction with traceable references.

Differentiate an Equation Without Solving for $y$