Differentiate an implicit relation

A point sliding on a circle

The orange point moves on the circle \(x^2 + y^2 = 13\), and the orange line is its tangent. Drag the point: the sum \(x^2 + y^2\) stays at 13, so its rate of change is 0.

At \((3, 2)\), a step of 0.01 to the right adds about \(2\cdot3\cdot0.01 = 0.06\) to \(x^2\). The sum must stay 13, so \(y^2\) loses about 0.06.

A small change in \(y\) changes \(y^2\) by about \(2y\) times that change, and \(2y = 4\) here, so \(y\) falls by about \(0.06 \div 4 = 0.015\). The slope is about \(-0.015 \div 0.01 = -1.5\).

The toggle draws a step along the tangent, much larger than 0.01 so that you can see it. The faint segment is the run to the right, and the rise is green when y grows and red when y falls.

y depends on x

This curve comes from an equation in \(x\) and \(y\), with no numbers shown. Near the orange point, it is the graph of a function \(y\) of \(x\), and the toggle shows that piece.

When \(x\) changes, \(y\) changes with it, and the slope \(\cy{\tfrac{dy}{dx}}\) of the orange tangent measures how fast.

A term that contains \(y\) changes because \(y\) changes. Its rate is its derivative with respect to \(y\) times \(\tfrac{dy}{dx}\), which is the chain rule with \(y\) as the inner function.

The equation stays true as the point moves, so its two sides change at the same rate.

Notation

1. Near the point, \(y\) is a function of \(x\), and its derivative is the slope of the tangent.

   \[y = y(x), \qquad \cy{\frac{dy}{dx}}\]

2. Differentiate both sides of the equation with respect to \(x\). A term in \(x\) only follows the usual rules.

   \[\frac{d}{dx}\,(x^2) = 2x\]

3. A term in \(y\) needs the chain rule, because \(y\) is a function of \(x\), the inner function.

   \[\frac{d}{dx}\,(y^3) = 3y^2\,\cy{\frac{dy}{dx}}\]

4. A product of \(x\) and \(y\) needs the product rule, because both factors change.

   \[\frac{d}{dx}\,(xy) = y + x\,\cy{\frac{dy}{dx}}\]

5. The equation holds along the curve, so the rate of the left side equals the rate of the right side.

   \[\frac{d}{dx}\bigl(\text{left side}\bigr) = \frac{d}{dx}\bigl(\text{right side}\bigr)\]

   This needs \(y\) to be a differentiable function of \(x\) near the point.

When it fails

At \((\sqrt{13}, 0)\), the tangent is vertical, and a vertical line has no slope. The dashed line just left of it meets the circle twice, so there the circle is not the graph of a function of \(x\).

Differentiating gives \(2x + 2y\,\tfrac{dy}{dx} = 0\), which at \((\sqrt{13}, 0)\) reads \(2\sqrt{13} = 0\). No value of \(\tfrac{dy}{dx}\) makes this true, so the method finds no slope here, as the red label shows.

Example 1: A term in y squared

Differentiate \(x^2 + 4y^2 = 20\) with respect to \(x\). Then find \(\tfrac{dy}{dx}\) at \((2, 2)\).

1. The term \(x^2\) has no \(y\), so its derivative is \(2x\).

   \[\frac{d}{dx}\,(x^2) = 2x\]

2. The term \(4y^2\) has \(y\) as the inner function, so the chain rule adds the factor \(\tfrac{dy}{dx}\).

   \[\frac{d}{dx}\,(4y^2) = 8y\,\cy{\frac{dy}{dx}}\]

3. The right side, 20, is a constant, so its derivative is 0.

   \[2x + 8y\,\cy{\frac{dy}{dx}} = 0\]

4. At \((2, 2)\), the equation reads \(4 + 16\,\tfrac{dy}{dx} = 0\), so \(\tfrac{dy}{dx} = -\tfrac14\). This is the slope of the orange tangent.

   \[\frac{dy}{dx}\bigg|_{(2,2)} = \cy{-\tfrac14}\]

Example 2: A product of x and y

Differentiate \(x^2 + xy + y^2 = 7\) with respect to \(x\). Then find \(\tfrac{dy}{dx}\) at \((2, 1)\).

1. The term \(x^2\) gives \(2x\). The term \(y^2\) needs the chain rule, and it gives \(2y\,\tfrac{dy}{dx}\).

   \[\begin{aligned} \frac{d}{dx}\,(x^2) &= 2x \\ \frac{d}{dx}\,(y^2) &= 2y\,\frac{dy}{dx} \end{aligned}\]

2. A tempting wrong move: The middle term is \(xy\). Differentiate \(x\) and leave \(y\) alone.

   \[\begin{aligned} \frac{d}{dx}\,(xy) &= y \\ 2x + y + 2y\,\frac{dy}{dx} &= 0 \end{aligned}\]

   At \((2, 1)\), this gives \(5 + 2\,\tfrac{dy}{dx} = 0\), so \(\tfrac{dy}{dx} = -\tfrac52\). The red line with this slope is steeper than the orange tangent, whose slope is \(-\tfrac54\). Both \(x\) and \(y\) change as the point moves, so \(xy\) changes in two ways.

3. Use the product rule, because both factors change.

   \[\frac{d}{dx}\,(xy) = y + x\,\cy{\frac{dy}{dx}}\]

4. Put the terms together. At \((2, 1)\), the equation reads \(5 + 4\,\tfrac{dy}{dx} = 0\), so \(\tfrac{dy}{dx} = -\tfrac54\), the slope of the orange tangent.

   \[\begin{aligned} 2x + y + x\,\frac{dy}{dx} + 2y\,\frac{dy}{dx} &= 0 \\ \frac{dy}{dx}\bigg|_{(2,1)} &= \cy{-\tfrac54} \end{aligned}\]

Practice

Differentiate \(xy^3 + 2x = 6\) with respect to \(x\). Then find \(\tfrac{dy}{dx}\) at \((2, 1)\).

Both factors of \(xy^3\) change as the point moves, so use the product rule. Then drag the point to \((2, 1)\) and compare the slope of the tangent with your answer.

Questions

Questions 1 and 2 use the hyperbola \(x^2 - y^2 = 5\). Its tangent at \((3, 2)\) has slope \(\tfrac32\), and a dashed vertical line is drawn at \(x = 3\).

1. At \((3, 2)\), what is the rate of change of \(y^2\) with respect to \(x\)?

   • 6
   • 4
   • 3
   Answer

   6

2. The dashed line meets the curve at two points. What does this show?

   • The whole curve is not one function of \(x\), but near each point it is.
   • The slope does not exist at \(x = 3\).
   • The method cannot be used on this curve.
   Answer

   The whole curve is not one function of \(x\), but near each point it is.

3. Differentiate \(x^2y^3\) with respect to \(x\).

   • \(2xy^3 + 3x^2y^2\,\dfrac{dy}{dx}\)
   • \(2xy^3\)
   • \(2x\cdot3y^2\,\dfrac{dy}{dx}\)
   • \(2xy^3 + 3x^2y^2\)
   Answer

   \(2xy^3 + 3x^2y^2\,\dfrac{dy}{dx}\)

Summary

You differentiate both sides with respect to \(x\), and you add the factor \(\tfrac{dy}{dx}\) to every term that contains \(y\).

This works because the equation holds along the curve, so both sides change at the same rate.

You use it when the equation is hard to solve for y, or when solving for y gives more than one function.

This is one of four skills in topic 3.2 of Which & Why: AP Calculus.