Wemazi

Topic 3.2

Solve for the derivative

Builds on differentiating an implicit relation. 7 minutes.

One unknown at each point

The orange point is at \((2, 1)\) on the curve \(x^2 + 2xy - y^2 = 7\). Differentiating this equation gives the equation on the figure, which holds at every point of the curve.

At \((2, 1)\), that equation becomes \(6 + 2\,\tfrac{dy}{dx} = 0\). Its one unknown is the slope, and its solution \(-3\) is the slope of the orange tangent.

Move the point to \((2, 3)\). The equation becomes \(10 - 2\,\tfrac{dy}{dx} = 0\), so the slope is 5 and the tangent rises steeply.

The numbers change from point to point, but the equation keeps its form: a number plus a number times the slope.

Any differentiated equation

After you differentiate, \(\tfrac{dy}{dx}\) appears in some terms and not in others. Call the terms without it \(B\), and write the terms with it as \(A\) times \(\tfrac{dy}{dx}\).

The equation reads \(B + A\,\tfrac{dy}{dx} = 0\). Drag the point. \(A\) and \(B\) change, because they depend on both \(x\) and \(y\), but at each point the equation has one unknown.

Its solution is \(\tfrac{dy}{dx} = -\tfrac{B}{A}\), a formula in \(x\) and \(y\) that works at every point where \(A\) is not 0.

Notation

  1. Start from the equation on the figure, with the bracket multiplied out.

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

  2. Keep the terms with \(\tfrac{dy}{dx}\) on the left, and move the other terms to the right.

    \[2x\,\frac{dy}{dx} - 2y\,\frac{dy}{dx} = -2x - 2y\]

  3. Factor \(\tfrac{dy}{dx}\) out of the left side. Then it appears in one place, so one division can isolate it.

    \[(2x - 2y)\,\frac{dy}{dx} = -2x - 2y\]

  4. Divide both sides by the bracket.

    \[\frac{dy}{dx} = \frac{-2x - 2y}{2x - 2y} = -\frac{x + y}{x - y}\]

  5. The formula contains both \(x\) and \(y\), so to find a slope you substitute both coordinates of a point on the curve.

    \[\left.\frac{dy}{dx}\right|_{(2,\,1)} = -\frac{3}{1} = \cy{-3}\]

When it fails

On the curve, \(x = y\) at \(\left(\sqrt{3.5}, \sqrt{3.5}\right)\), about \((1.87, 1.87)\), and there the bracket \(2x - 2y\) is 0.

The equation then reads \(7.48 + 0\cdot\tfrac{dy}{dx} = 0\), which no number satisfies, so the slope does not exist and the tangent is vertical. The formula holds only where \(x \neq y\).

Example 1: Collect and factor

Find \(\tfrac{dy}{dx}\) for \(y^3 + 2xy = 12\). Then find the slope at \((1, 2)\).

  1. Differentiate both sides. The term \(y^3\) needs the chain rule, and the term \(2xy\) needs the product rule.

    \[3y^2\,\cy{\frac{dy}{dx}} + 2y + 2x\,\cy{\frac{dy}{dx}} = 0\]

  2. Move the term without \(\tfrac{dy}{dx}\) to the right.

    \[3y^2\,\frac{dy}{dx} + 2x\,\frac{dy}{dx} = -2y\]

  3. Factor \(\tfrac{dy}{dx}\) out of the left side.

    \[(3y^2 + 2x)\,\frac{dy}{dx} = -2y\]

  4. Divide by the bracket.

    \[\frac{dy}{dx} = -\frac{2y}{3y^2 + 2x}\]

  5. The point \((1, 2)\) is on the curve, because \(8 + 4 = 12\). There, the slope is \(-\tfrac{4}{14}\).

    \[\frac{dy}{dx}\bigg|_{(1,2)} = \cy{-\tfrac27}\]

Example 2: A term that is easy to leave out

Find \(\tfrac{dy}{dx}\) for \(x^2y + y^2 = 12\). Then find the slope at \((2, 2)\).

  1. Differentiate both sides. The product \(x^2y\) gives two terms.

    \[2xy + x^2\,\frac{dy}{dx} + 2y\,\frac{dy}{dx} = 0\]

  2. A tempting wrong move: Leave out \(x^2\,\tfrac{dy}{dx}\), which looks like a term in \(x\), and solve with \(2y\,\tfrac{dy}{dx}\) alone.

    \[\begin{aligned} 2y\,\frac{dy}{dx} &= -2xy \\ \frac{dy}{dx} &= -x \end{aligned}\]

    At \((2, 2)\), this gives slope \(-2\). The red line with slope \(-2\) cuts across the curve, but the orange tangent has slope \(-1\). The term \(x^2\,\tfrac{dy}{dx}\) also contains the unknown, so it must stay with \(2y\,\tfrac{dy}{dx}\).

  3. Move \(2xy\) to the right, and factor \(\tfrac{dy}{dx}\) out of both terms on the left.

    \[(x^2 + 2y)\,\frac{dy}{dx} = -2xy\]

  4. Divide by the bracket. At \((2, 2)\), the slope is \(-\tfrac88 = -1\), which matches the orange tangent.

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

Practice

Find \(\tfrac{dy}{dx}\) for \(xy^2 + x^2 = 10\). Then find the slope at \((1, 3)\).

Drag the point to \((1, 3)\). The tangent falls steeply, so the slope is less than \(-1\).

Solution

  1. The product \(xy^2\) gives \(y^2 + 2xy\,\tfrac{dy}{dx}\), and the term \(x^2\) gives \(2x\).

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

  2. Move the terms without \(\tfrac{dy}{dx}\) to the right, and divide by \(2xy\).

    \[\frac{dy}{dx} = -\frac{y^2 + 2x}{2xy}\]

  3. The point \((1, 3)\) is on the curve, because \(9 + 1 = 10\). There, the slope is \(-\tfrac{11}{6}\).

    \[\frac{dy}{dx}\bigg|_{(1,3)} = \cy{-\tfrac{11}{6}}\]

Questions

Questions 1 and 2 use the curve \(xy + y^2 = 6\). It passes through \(A = (1, 2)\) and \(B = (5, 1)\), and its slope is \(\tfrac{dy}{dx} = -\tfrac{y}{x + 2y}\).

  1. What slope does the orange tangent have at \(A = (1, 2)\)?

    • \(-\tfrac25\)
    • \(\tfrac25\)
    • \(-\tfrac52\)
    Answer

    \(-\tfrac25\)

  2. How does the tangent at \(B = (5, 1)\) compare with the tangent at \(A\)?

    • It falls more gently.
    • It is steeper.
    • It rises.
    Answer

    It falls more gently.

  3. Solve \(2x + 3y^2\,\dfrac{dy}{dx} = 4\,\dfrac{dy}{dx}\) for \(\dfrac{dy}{dx}\).

    • \(\dfrac{2x}{4 - 3y^2}\)
    • \(-\dfrac{2x}{3y^2 + 4}\)
    • \(-\dfrac{2x}{3y^2}\)
    Answer

    \(\dfrac{2x}{4 - 3y^2}\)

Summary

You move every term with \(\tfrac{dy}{dx}\) to one side, factor \(\tfrac{dy}{dx}\) out, and divide by the bracket.

This works because \(\tfrac{dy}{dx}\) appears only as a factor in some terms, so after you factor, it appears once and one division isolates it.

The result is a formula in \(x\) and \(y\), so to find a slope you substitute both coordinates of a point on the curve.

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