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Amy Ferguson Moncure

Calculus BC ยท Unit 3

Implicit Differentiation

Differentiate both sides, solve for dy/dx, and evaluate on the curve.

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Key Ideas Easy Medium Hard

\(\dfrac{d}{dx}\bigl[y\bigr] = \dfrac{dy}{dx}\)

What to Remember

Differentiate both sides with respect to x

Every time you differentiate a \(y\) term, multiply by \(\dfrac{dy}{dx}\). A product \(xy\) still needs the product rule.

Then solve for the derivative

Collect every term that contains \(dy/dx\) on one side. Factor \(dy/dx\), then divide.

Substitute the point after you solve

Find \(dy/dx\) as an expression in \(x\) and \(y\). Put the coordinates in on the next line. Both coordinates belong in that line when both appear.

A second derivative needs the first one again

Differentiate \(dy/dx\) with respect to \(x\). Replace \(dy/dx\) with the expression you already found, then substitute the point.

Level 1

Easy

The relation is a circle, a product, or a power. Solve for dy/dx, then use the point.

Example
Find \(\dfrac{dy}{dx}\) at \((3, 4)\) on \(x^2 + y^2 = 25\).

\(x^2 + y^2 = 25\)

\(2x + 2y\,\dfrac{dy}{dx} = 0\)

\(\dfrac{dy}{dx} = -\dfrac{x}{y}\)

Substitute \((3, 4)\).

\(-\dfrac{3}{4}\)

Answer: \(-\dfrac{3}{4}\)

1
Find \(\dfrac{dy}{dx}\) at \((6, 8)\) on \(x^2 + y^2 = 100\).
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\(2x + 2y\,\dfrac{dy}{dx} = 0\)

\(\dfrac{dy}{dx} = -\dfrac{x}{y}\)

\(-\dfrac{6}{8}\)

\(-\dfrac{3}{4}\)

Answer: \(-\dfrac{3}{4}\)

2
Find \(\dfrac{dy}{dx}\) at \((3, 4)\) on \(xy = 12\).
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\(x\,\dfrac{dy}{dx} + y = 0\)

\(\dfrac{dy}{dx} = -\dfrac{y}{x}\)

\(-\dfrac{4}{3}\)

Answer: \(-\dfrac{4}{3}\)

3
Find \(\dfrac{dy}{dx}\) at \((1, 2)\) on \(x + y^3 = 9\).
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\(1 + 3y^2\,\dfrac{dy}{dx} = 0\)

\(\dfrac{dy}{dx} = -\dfrac{1}{3y^2}\)

\(-\dfrac{1}{3\cdot 4}\)

\(-\dfrac{1}{12}\)

Answer: \(-\dfrac{1}{12}\)

4
Find \(\dfrac{dy}{dx}\) at \((1, 2)\) on \(y^2 = 4x\).
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\(2y\,\dfrac{dy}{dx} = 4\)

\(\dfrac{dy}{dx} = \dfrac{2}{y}\)

\(\dfrac{2}{2}\)

\(1\)

Answer: \(1\)

5
Find \(\dfrac{dy}{dx}\) at \((5, 4)\) on \(x^2 - y^2 = 9\).
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\(2x - 2y\,\dfrac{dy}{dx} = 0\)

\(\dfrac{dy}{dx} = \dfrac{x}{y}\)

\(\dfrac{5}{4}\)

Answer: \(\dfrac{5}{4}\)

Level 2

Medium

A product of x and y, or a function of y, still gets a factor of dy/dx.

Example
Find \(\dfrac{dy}{dx}\) at \((1, 2)\) on \(x^2 y + y^2 = 6\).

\(2xy + x^2\,\dfrac{dy}{dx} + 2y\,\dfrac{dy}{dx} = 0\)

\(\dfrac{dy}{dx}\bigl(x^2 + 2y\bigr) = -2xy\)

\(\dfrac{dy}{dx} = -\dfrac{2xy}{x^2 + 2y}\)

Substitute \((1, 2)\).

\(-\dfrac{4}{1 + 4}\)

\(-\dfrac{4}{5}\)

Answer: \(-\dfrac{4}{5}\)

1
Find \(\dfrac{dy}{dx}\) at \((1, 0)\) on \(x + \sin y = 1\).
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\(1 + \cos y\,\dfrac{dy}{dx} = 0\)

\(\dfrac{dy}{dx} = -\dfrac{1}{\cos y}\)

\(-\dfrac{1}{\cos 0}\)

\(-1\)

Answer: \(-1\)

2
Find \(\dfrac{dy}{dx}\) at \((2, 0)\) on \(e^{y} + x^2 = 5\).
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\(e^{y}\,\dfrac{dy}{dx} + 2x = 0\)

\(\dfrac{dy}{dx} = -2x\, e^{-y}\)

\(-4\, e^{0}\)

\(-4\)

Answer: \(-4\)

3
Find \(\dfrac{dy}{dx}\) at \((1, 2)\) on \(x^2 + xy + y^2 = 7\).
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\(2x + y + x\,\dfrac{dy}{dx} + 2y\,\dfrac{dy}{dx} = 0\)

\(\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}\)

Substitute \((1, 2)\).

\(-\dfrac{4}{5}\)

Answer: \(-\dfrac{4}{5}\)

4
Find \(\dfrac{dy}{dx}\) at \((1, 1)\) on \(y + \ln y = x\).
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\(\dfrac{dy}{dx} + \dfrac{1}{y}\,\dfrac{dy}{dx} = 1\)

\(\dfrac{dy}{dx}\left(1 + \dfrac{1}{y}\right) = 1\)

\(\dfrac{dy}{dx} = \dfrac{y}{y + 1}\)

\(\dfrac{1}{2}\)

Answer: \(\dfrac{1}{2}\)

5
Which expression equals \(\dfrac{dy}{dx}\) on \(x^3 + y^3 = 9\)?
A. \(-\dfrac{x^2}{y^2}\)
B. \(-\dfrac{y^2}{x^2}\)
C. \(-3x^2\)
D. \(\dfrac{x^2}{y^2}\)
Show Answer

\(3x^2 + 3y^2\,\dfrac{dy}{dx} = 0\)

\(\dfrac{dy}{dx} = -\dfrac{x^2}{y^2}\)

Answer: A. \(-\dfrac{x^2}{y^2}\)

Level 3

Hard

A tangent line, or a second derivative at the point.

Example
Find the tangent line to \(x^2 + y^2 = 25\) at \((3, -4)\).

\(\dfrac{dy}{dx} = -\dfrac{x}{y}\)

Substitute \((3, -4)\).

\(-\dfrac{3}{-4} = \dfrac{3}{4}\)

\(y - (-4) = \dfrac{3}{4}(x - 3)\)

\(y + 4 = \dfrac{3}{4}(x - 3)\)

Answer: \(y + 4 = \dfrac{3}{4}(x - 3)\)

1
Find \(\dfrac{dy}{dx}\) at \((1, 1)\) on \(xy + y^2 = 2x\).
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\(y + x\,\dfrac{dy}{dx} + 2y\,\dfrac{dy}{dx} = 2\)

Substitute \((1, 1)\).

\(1 + \dfrac{dy}{dx} + 2\,\dfrac{dy}{dx} = 2\)

\(1 + 3\,\dfrac{dy}{dx} = 2\)

\(\dfrac{dy}{dx} = \dfrac{1}{3}\)

Answer: \(\dfrac{1}{3}\)

2
Find \(\dfrac{dy}{dx}\) at \((-1, 1)\) on \(\cos(x + y) = y\).
Show Answer

\(-\sin(x + y)\left(1 + \dfrac{dy}{dx}\right) = \dfrac{dy}{dx}\)

At \((-1, 1)\), \(x + y = 0\) and \(\sin 0 = 0\).

\(0 = \dfrac{dy}{dx}\)

Answer: \(0\)

3
Find \(\dfrac{dy}{dx}\) at \((2, 1)\) on \(x^2 y = 4\).
Show Answer

\(2xy + x^2\,\dfrac{dy}{dx} = 0\)

\(\dfrac{dy}{dx} = -\dfrac{2y}{x}\)

\(-\dfrac{2}{2}\)

\(-1\)

Answer: \(-1\)

4
On \(xy = 6\), find \(\dfrac{d^2 y}{dx^2}\) at \((2, 3)\).
Show Answer

\(\dfrac{dy}{dx} = -\dfrac{y}{x}\)

\(\dfrac{d^2 y}{dx^2} = -\dfrac{x\,\dfrac{dy}{dx} - y}{x^2}\)

Substitute \(\dfrac{dy}{dx} = -\dfrac{3}{2}\) and \((2, 3)\).

\(2\left(-\dfrac{3}{2}\right) - 3\)

\(-6\)

\(-\dfrac{-6}{4}\)

\(\dfrac{3}{2}\)

Answer: \(\dfrac{3}{2}\)

5
On \(x^2 + y^2 = 25\), find \(\dfrac{d^2 y}{dx^2}\) at \((3, 4)\).
Show Answer

\(\dfrac{dy}{dx} = -\dfrac{x}{y}\)

\(\dfrac{d^2 y}{dx^2} = \dfrac{-y + x\,\dfrac{dy}{dx}}{y^2}\)

Substitute \(\dfrac{dy}{dx} = -\dfrac{3}{4}\) and \((3, 4)\).

\(-4 + 3\left(-\dfrac{3}{4}\right)\)

\(-\dfrac{25}{4}\)

\(\dfrac{-\dfrac{25}{4}}{16}\)

\(-\dfrac{25}{64}\)

Answer: \(-\dfrac{25}{64}\)

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