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Calculus 1

Implicit differentiation

An equation such as x^3 + y^3 = 6xy draws a smooth curve, but it cannot be solved neatly for y. How can we find the slope of its tangent lines anyway?

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Second and higher derivativesCalculus 1The chain ruleCalculus 1Tangent lines and normal linesCalculus 1
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Implicit differentiation
Derivatives of inverse functionsDerivatives of inverse trigonometric functions

Why this exists

Every derivative so far began with a formula y = f(x). Many curves arrive instead as one equation in x and y together. The circle x^2 + y^2 = 225 fails the Testing graphs and equations for functions, so no single formula y = f(x) describes it. The curve x^3 + y^3 = 6xy in Figure 2 is worse: solving it for y means solving a cubic equation for each x. Yet both curves are smooth, and zooming in on any ordinary point shows something that looks like a line. Implicit differentiation finds the slope of that line without ever solving for y.

Day 1 sets up the method and finds tangent lines. Day 2 adds horizontal and vertical tangents, second derivatives, the power rule for rational exponents, and the method’s limits.

Definition

Day 1: y as an unnamed function of x

An equation in x and y defines y implicitly as a function of x near a point (a, b) of its graph when there is a differentiable function f on an open interval around a with f(a) = b, such that near (a, b) the graph of the equation is exactly the graph of y = f(x). Near (9, 12) the circle above is the graph of \displaystyle y = \sqrt{225 - x^2}, and near (9, -12) of \displaystyle y = -\sqrt{225 - x^2}. Near (15, 0) it is the graph of no function of x at all, since every vertical line just left of x = 15 meets it twice.

Implicit differentiation finds \dfrac{dy}{dx} without a formula for f: keep writing y for the unknown f(x), differentiate both sides of the equation with respect to x, and solve the result for \dfrac{dy}{dx}. Terms in x alone are differentiated as usual. A term built from y is a function of the function f(x), so the chain rule attaches a factor \dfrac{dy}{dx}:

\frac{d}{dx}\big[y^n\big] = n\,y^{n-1}\,\frac{dy}{dx}

In the same way \dfrac{d}{dx}\big[\sin y\big] = \cos y\,\dfrac{dy}{dx}, and a product of x and y takes the product rule: \dfrac{d}{dx}\big[xy\big] = y + x\,\dfrac{dy}{dx}.

The result usually contains both x and y, as it must: one vertical line can meet the curve at several points with different slopes. The slope at a particular point (a, b) of the curve is written \dfrac{dy}{dx}\Big|_{(a,\,b)}. We also write y' for \dfrac{dy}{dx}.

Day 2: special tangents and the second derivative

Suppose implicit differentiation gives \dfrac{dy}{dx} = \dfrac{N}{D}, where N and D are expressions in x and y. A curve has a horizontal tangent at a point where its slope is 0. It has a vertical tangent at a point where, nearby, the curve is the graph of x as a differentiable function of y whose derivative \dfrac{dx}{dy} is 0 there. The second derivative \dfrac{d^2y}{dx^2} is the derivative of the expression for \dfrac{dy}{dx}, found the same way, with y still standing for a function of x.

Intuition

Think of the equation as a balance that holds at every point of the curve. As a point slides along the curve, both sides change but stay equal, so they change at the same rate. Differentiating both sides writes down exactly that, and each y changes at the rate \dfrac{dy}{dx}, which is why that factor appears every time a y is differentiated.

x
y
(9, 12)
(9, -12)
Figure 1The circle of Example 1. Near (9, 12) it is the graph of the top half, near (9, -12) of the bottom half, and the one formula -\dfrac{x}{y} gives both slopes.

In Figure 1 one formula serves both halves of the circle; substituting the point picks the half.

Results

Implicit differentiation. If y = f(x) is differentiable and satisfies an equation for every x in an open interval, then the two sides of the equation have equal derivatives with respect to x there, where each term in y is differentiated by the chain rule:

\frac{d}{dx}\big[g(y)\big] = g'(y)\,\frac{dy}{dx}

Tangent line. At (a, b) on the curve, the tangent line is y - b = m(x - a) with m = \dfrac{dy}{dx}\Big|_{(a,\,b)}.

Horizontal and vertical tangents. If \dfrac{dy}{dx} = \dfrac{N}{D} and (a, b) lies on the curve: when N(a, b) = 0 and D(a, b) \ne 0 the tangent is horizontal; when D(a, b) = 0 and N(a, b) \ne 0 the tangent is vertical; when both are 0 the formula decides nothing.

The power rule for rational exponents. For integers p and q with q \ge 1 and \displaystyle \frac{p}{q} in lowest terms,

\frac{d}{dx}\big[x^{p/q}\big] = \frac{p}{q}\,x^{p/q - 1}

for every x > 0, and for every x \ne 0 when q is odd.

Why it is true

Differentiating both sides. Suppose y = f(x) satisfies the equation for every x in an open interval. With f(x) written in for y, the left side and the right side are then the same function of x on that interval, since they agree at every x. Equal functions have equal derivatives. The chain rule gives \dfrac{d}{dx}\big[g(f(x))\big] = g'(f(x))\,f'(x), which is g'(y)\,\dfrac{dy}{dx}. The argument assumes that a differentiable f exists; it computes \dfrac{dy}{dx}, and it does not prove that there is one (see Boundaries).

Vertical tangents, an informal argument. In the differentiated equation, the factor \dfrac{dy}{dx} appears only where a y was differentiated, once each time. Collecting, the equation reads A + B\,\dfrac{dy}{dx} = 0, so \dfrac{dy}{dx} = -\dfrac{A}{B}, with N = -A and D = B. Now treat x as a function of y and differentiate with respect to y. The same pieces appear, but the factor \dfrac{dx}{dy} now lands on the differentiated x parts, giving A\,\dfrac{dx}{dy} + B = 0, and

\frac{dx}{dy} = -\frac{B}{A} = \frac{D}{N}

Where D = 0 and N \ne 0, this gives \dfrac{dx}{dy} = 0: the curve, seen as a graph of x against y, has a level tangent, which in the xy-plane is a vertical tangent. The argument assumes that x is a differentiable function of y near the point; the implicit function theorem (Boundaries), with x and y swapped, supplies that when N \ne 0.

The power rule for \displaystyle \frac{p}{q}. Let u = x^{1/q}, the q-th root, defined for x > 0 and, when q is odd, for x < 0 too. The power rule and basic differentiation rules proved that x^{1/q} is differentiable for x > 0 by factoring a^q - b^q, and the same factoring works for x < 0 when q is odd. So y = x^{p/q} = u^p is differentiable wherever x \ne 0, by the chain rule and the power rule for integer exponents. Raising y = u^p to the power q gives y^q = u^{pq} = x^p. Differentiating both sides with respect to x, using the integer power rule on each side and the chain rule on the left,

q\,y^{q-1}\,\frac{dy}{dx} = p\,x^{p-1}

Since x \ne 0, y \ne 0, so we may divide. With x = u^q and y = u^p, every power below is an integer power of the nonzero number u:

\begin{aligned} \frac{dy}{dx} &= \frac{p}{q} \cdot \frac{x^{p-1}}{y^{q-1}} = \frac{p}{q} \cdot \frac{u^{q(p-1)}}{u^{p(q-1)}} \\ &= \frac{p}{q}\,u^{p - q} = \frac{p}{q}\,x^{p/q - 1} \end{aligned}

since u^{p-q} = \big(x^{1/q}\big)^{p-q}. This is the proof The power rule and basic differentiation rules said would come here.

Worked examples

Example 1

Day 1: the circle, checked against its two halves

Find \dfrac{dy}{dx} for the circle x^2 + y^2 = 225 by implicit differentiation, find the slopes at (9, 12) and (9, -12), and check both against the explicit formulas for the two halves.

Solution

Differentiate both sides with respect to x. The term x^2 gives 2x; the term y^2 gives 2y\,\dfrac{dy}{dx} by the chain rule; the constant 225 gives 0:

2x + 2y\,\frac{dy}{dx} = 0

Subtracting 2x and dividing by 2y (allowed where y \ne 0) gives \dfrac{dy}{dx} = -\dfrac{x}{y}. At (9, 12) the slope is -\dfrac{9}{12} = -\dfrac{3}{4}, and at (9, -12) it is \dfrac{3}{4}.

To check, the top half is y = (225 - x^2)^{1/2}, and by the chain rule

\frac{dy}{dx} = \frac{1}{2}(225 - x^2)^{-1/2}(-2x) = -\frac{x}{\sqrt{225 - x^2}}

which at x = 9 is \displaystyle -\dfrac{9}{\sqrt{144}} = -\dfrac{3}{4}. The bottom half \displaystyle y = -\sqrt{225 - x^2} has the opposite derivative, \displaystyle \dfrac{x}{\sqrt{225 - x^2}}, which is \dfrac{3}{4} at x = 9. Both explicit answers are -\dfrac{x}{y} with each half’s own y, so the implicit formula covers the whole circle at once. The tangent slope -\dfrac{x}{y} is also the negative reciprocal of the radius slope \dfrac{y}{x}, as Proving a tangent is perpendicular to the radius requires.

Example 2

Day 1: a product term and a tangent line

Find the tangent line to the curve x^2 y + y^3 = 10 at the point (3, 1).

Solution

First, the point is on the curve: 9 \cdot 1 + 1 = 10. The term x^2 y is a product of two functions of x, so the product rule gives 2xy + x^2\,\dfrac{dy}{dx}. The term y^3 gives 3y^2\,\dfrac{dy}{dx}, and the constant gives 0:

2xy + x^2\,\frac{dy}{dx} + 3y^2\,\frac{dy}{dx} = 0

Collect the terms with \dfrac{dy}{dx} and factor it out: (x^2 + 3y^2)\,\dfrac{dy}{dx} = -2xy, so

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

At (3, 1) the slope is -\dfrac{6}{9 + 3} = -\dfrac{1}{2}, and the tangent line is y - 1 = -\dfrac{1}{2}(x - 3), that is, y = -\dfrac{1}{2}x + \dfrac{5}{2}.

The denominator x^2 + 3y^2 is 0 only at the origin, which is not on the curve, so every point of the curve has a slope.

Example 3

Day 1: a tangent line to the folium

The curve x^3 + y^3 = 6xy is called the folium of Descartes (Figure 2). Find its tangent line at \displaystyle \left(\frac{4}{3}, \frac{8}{3}\right).

Solution

The point is on the curve: \displaystyle \frac{64}{27} + \frac{512}{27} = \frac{576}{27} = \frac{64}{3}, and \displaystyle 6 \cdot \frac{4}{3} \cdot \frac{8}{3} = \frac{64}{3}. Differentiate both sides, using the product rule on 6xy:

3x^2 + 3y^2\,\frac{dy}{dx} = 6y + 6x\,\frac{dy}{dx}

Move the \dfrac{dy}{dx} terms to the left and the rest to the right: (3y^2 - 6x)\,\dfrac{dy}{dx} = 6y - 3x^2. Dividing by 3 and then by y^2 - 2x,

\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}

At the point, the top is \displaystyle \frac{16}{3} - \frac{16}{9} = \frac{32}{9} and the bottom is \displaystyle \frac{64}{9} - \frac{24}{9} = \frac{40}{9}, so the slope is \dfrac{32}{40} = \dfrac{4}{5}. The tangent line is \displaystyle y - \frac{8}{3} = \frac{4}{5}\left(x - \frac{4}{3}\right), which simplifies to

y = \frac{4}{5}x + \frac{8}{5}
x
y
2
4
2
4
A
H
V
Figure 2The folium x^3 + y^3 = 6xy. The tangent at \displaystyle A = \left(\frac{4}{3}, \frac{8}{3}\right) (Example 3), the horizontal tangent at H and the vertical tangent at V (Example 4). At the origin two pieces of the curve cross.
Example 4

Day 2: horizontal and vertical tangents on the folium

Find every point of the folium x^3 + y^3 = 6xy where the tangent is horizontal or vertical, and say what the slope formula gives at the origin.

Solution

From Example 3, N = 2y - x^2 and D = y^2 - 2x. A horizontal tangent needs N = 0 at a point of the curve, so \displaystyle y = \frac{x^2}{2} and the curve’s equation must hold together. Substituting,

x^3 + \frac{x^6}{8} = 6x \cdot \frac{x^2}{2} = 3x^3

so \displaystyle \frac{x^6}{8} = 2x^3, that is, x^3(x^3 - 16) = 0. The root x = 0 gives the origin. The root x^3 = 16 gives \displaystyle x = \sqrt[3]{16} = 2\sqrt[3]{2} and \displaystyle y = \frac{x^2}{2} = 2\sqrt[3]{4}. There \displaystyle D = 8\sqrt[3]{2} - 4\sqrt[3]{2} = 4\sqrt[3]{2} \ne 0, so the tangent at \displaystyle H = \big(2\sqrt[3]{2},\, 2\sqrt[3]{4}\big) \approx (2.52, 3.17) is horizontal.

Exchanging x and y leaves the equation unchanged, so the curve is symmetric across y = x, and the reflection swaps N = 0 and D = 0. The same algebra with the letters swapped gives D = 0 and N \ne 0 at \displaystyle V = \big(2\sqrt[3]{4},\, 2\sqrt[3]{2}\big) \approx (3.17, 2.52), a vertical tangent, plus the origin again.

At the origin N = 0 and D = 0, so the formula reads \dfrac{0}{0} and says nothing. Figure 2 shows why: two pieces of the curve cross there, one leaving along each axis, so near the origin the curve is not the graph of any single function of x.

Example 5

Day 2: a second derivative, using the curve’s own equation

For the hyperbola x^2 - 4y^2 = 9, find \dfrac{d^2y}{dx^2} in terms of y alone and its value at (5, 2). Check the value with an explicit formula.

Solution

Differentiating once, 2x - 8y\,y' = 0, so y' = \dfrac{x}{4y}. Differentiate this quotient again by the quotient rule, remembering that y is a function of x, so the derivative of 4y is 4y':

y'' = \frac{4y \cdot 1 - x \cdot 4y'}{16y^2} = \frac{y - x\,y'}{4y^2}

The answer should not contain y', so substitute y' = \dfrac{x}{4y}. The top becomes y - \dfrac{x^2}{4y} = \dfrac{4y^2 - x^2}{4y}, and

y'' = \frac{4y^2 - x^2}{16y^3}

The curve’s own equation says 4y^2 - x^2 = -9 at every point of the curve, so

\frac{d^2y}{dx^2} = -\frac{9}{16y^3}

The point (5, 2) is on the curve (25 - 16 = 9), and there y' = \dfrac{5}{8} and y'' = -\dfrac{9}{128}.

Check: near (5, 2) the curve is \displaystyle y = \frac{1}{2}(x^2 - 9)^{1/2}, so \displaystyle y' = \dfrac{x}{2\sqrt{x^2 - 9}}, which is \dfrac{5}{8} at x = 5. By the product rule on \displaystyle \frac{1}{2}x(x^2 - 9)^{-1/2},

\begin{aligned} y'' &= \tfrac{1}{2}\Big[(x^2 - 9)^{-1/2} - x^2 (x^2 - 9)^{-3/2}\Big] \\ &= -\frac{9}{2(x^2 - 9)^{3/2}} \end{aligned}

which at x = 5 is -\dfrac{9}{2 \cdot 64} = -\dfrac{9}{128}. The sign of -\dfrac{9}{16y^3} says the upper halves of the hyperbola bend downward and the lower halves upward, as Concavity and inflection points will make precise.

Example 6

Day 2: a rational exponent by implicit differentiation

Let y = x^{3/5}. (a) Find \dfrac{dy}{dx} by differentiating y^5 = x^3 implicitly, and compare with the power rule. (b) Find the tangent line at x = 32. (c) Find the slope at x = -32, and say what happens at x = 0.

Solution

(a) Raising both sides to the fifth power gives y^5 = x^3, true for every real x, since the fifth root of a negative number is negative. Differentiating, 5y^4\,y' = 3x^2, so y' = \dfrac{3x^2}{5y^4} wherever y \ne 0. Since y^4 = x^{12/5}, this is \dfrac{3}{5}x^{2 - 12/5} = \dfrac{3}{5}x^{-2/5}, the power rule’s answer.

(b) At x = 32, \displaystyle y = \big(\sqrt[5]{32}\big)^3 = 8, so y' = \dfrac{3 \cdot 1024}{5 \cdot 4096} = \dfrac{3}{20}. The tangent line is y - 8 = \dfrac{3}{20}(x - 32), that is, y = \dfrac{3}{20}x + \dfrac{16}{5}.

(c) At x = -32, y = -8, and x^2 and y^4 are the same as before, so the slope is again \dfrac{3}{20}, as the symmetry of this odd function’s graph about the origin predicts. At x = 0 both 3x^2 and 5y^4 are 0, and x^{-2/5} is undefined. The difference quotient \dfrac{h^{3/5}}{h} = h^{-2/5} grows without bound as h \to 0, so the graph has a vertical tangent there, one of the shapes in Differentiability: corners, cusps and piecewise functions.

Example 7

Day 2: where the formula means nothing

The curve y^2 = x^2(x + 4) is drawn in Figure 3. (a) Find \dfrac{dy}{dx} and its value at (-3, 3). (b) Breesha substitutes (1, 1) into the formula. What does her number mean? (c) What happens at the origin? Find the slopes of the two pieces \displaystyle y = x\sqrt{x + 4} and \displaystyle y = -x\sqrt{x + 4} there.

Solution

(a) The right side is x^3 + 4x^2, so differentiating gives 2y\,y' = 3x^2 + 8x and

\frac{dy}{dx} = \frac{3x^2 + 8x}{2y}

At (-3, 3), on the curve since 9 = 9 \cdot 1, the slope is \dfrac{27 - 24}{6} = \dfrac{1}{2}.

(b) At (1, 1) the formula gives \dfrac{11}{2}, but (1, 1) is not on the curve: 1 \ne 1 \cdot 5. The formula was derived by assuming the point satisfies the equation, so this number is the slope of nothing. The points of the curve with x = 1 are \displaystyle \big(1, \pm\sqrt{5}\big), with slopes \displaystyle \pm\dfrac{11}{2\sqrt{5}}.

(c) At the origin the formula reads \dfrac{0}{0}. Taking square roots, the curve consists of the two graphs \displaystyle y = x\sqrt{x + 4} and \displaystyle y = -x\sqrt{x + 4} for x \ge -4. By the product rule the first has derivative \displaystyle \sqrt{x + 4} + \dfrac{x}{2\sqrt{x + 4}}, which is 2 at x = 0, and the second has derivative -2 there. Two pieces cross at the origin with different slopes, so near it the curve is the graph of no single function of x, and \dfrac{dy}{dx} has no meaning there. At (-4, 0), by contrast, N = 16 \ne 0 and D = 0: a vertical tangent.

x
y
-2
1
2
-2
(-3, 3)
Figure 3The curve y^2 = x^2(x + 4) of Example 7. At the origin two pieces cross with slopes 2 and -2, so no single slope exists there.

Boundaries

  • The point must lie on the curve. The formula for \dfrac{dy}{dx} accepts any x and y, but it is a slope only at points that satisfy the original equation (Example 7(b)).
  • The method assumes y is a differentiable function of x near the point. It computes \dfrac{dy}{dx} when there is one; it does not show that there is one. The implicit function theorem, proved in multivariable calculus and stated here without proof, supplies the guarantee: if the equation is built from functions with continuous derivatives, and the coefficient D of \dfrac{dy}{dx} (before any canceling) is not 0 at a point of the curve, then near that point the curve is the graph of a differentiable function of x, and its slope is \dfrac{N}{D}.
  • Where D = 0, many things can happen. A vertical tangent (the circle at (15, 0)), two pieces crossing (the folium and Example 7 at the origin), a cusp (exercise 21), or a single isolated point (x^2 + y^2 = 0 is the origin alone).
  • When N = 0 and D = 0 together, the formula decides nothing, neither horizontal nor vertical; the curve near the point must be studied another way.
  • The power rule for \displaystyle \frac{p}{q} needs x > 0, or x \ne 0 when q is odd. At x = 0 the derivative may fail to exist, as for x^{3/5} in Example 6.

Where this goes next

Differentiating both sides of an equation with respect to time instead of x turns a fixed relationship between quantities into a relationship between their rates, which is related rates. Writing y = \ln x as e^y = x and differentiating implicitly gives the derivative of the logarithm in Derivatives of logarithms and logarithmic differentiation. The same move on f(y) = x gives the derivative of an inverse function and the derivatives of the inverse trigonometric functions.

Exercises

Day 1: finding dy/dx

  1. For x^2 + 3y^2 = 21, find \dfrac{dy}{dx} and its value at (3, 2).
  2. For y^3 + y = x^2, find \dfrac{dy}{dx}.
  3. For x^3 + xy^2 = 2y, find \dfrac{dy}{dx}.
  4. For \sin y + xy = x^2, find \dfrac{dy}{dx}.
  5. For (x + y)^3 = 8x, find \dfrac{dy}{dx}.

Day 1: tangent lines

  1. Find the tangent line to x^2 + xy + y^2 = 7 at (1, 2).
  2. Check that (5, 2) lies on y^4 - xy = 6, and find the tangent line there.
  3. The curve (x^2 + y^2)^2 = 4xy is a lemniscate, a figure-eight through the origin. Find the tangent line at (1, 1), and explain why the slope fits the curve’s symmetry across the line y = x.
  4. Find the tangent line and the normal line to x^3 + y^3 = 9 at (2, 1).

Day 2: horizontal and vertical tangents

  1. Find every point of x^2 - xy + y^2 = 12 where the tangent is horizontal, and every point where it is vertical.
  2. For y^2 = x^3 - 4x, find every point with a horizontal tangent and every point with a vertical tangent. Explain why \displaystyle x = \frac{2}{\sqrt{3}} gives no point, although it makes the top of the slope formula 0.
  3. Find the points of x^2 + 4y^2 = 32 where the tangent is parallel to the line x + 2y = 0.

Day 2: second derivatives

  1. For x^3 + y^3 = 9, show that \dfrac{d^2y}{dx^2} = -\dfrac{18x}{y^5}, and find its value at (2, 1).
  2. For y^2 = 12x, find \dfrac{d^2y}{dx^2} in terms of y, and its value at (3, 6).
  3. For xy + y^2 = 4, show that \dfrac{d^2y}{dx^2} = \dfrac{8}{(x + 2y)^3}, and find its value at (3, 1).

Day 2: rational exponents

  1. Find f'(32) for f(x) = x^{2/5}, and g'(4) for \displaystyle g(x) = \dfrac{1}{\sqrt{x}}.
  2. Find the derivative of y = x^{2/3} by differentiating y^3 = x^2 implicitly, not by the power rule. Use it to find the slope at x = -27, and check with the power rule.
  3. Find the tangent line to \displaystyle y = \sqrt[3]{x} at x = 8, and say what happens at x = 0.

Reasoning

  1. Illiam differentiates x^2 + y^2 = 25, gets -\dfrac{x}{y}, and substitutes (2, 4) to get -\dfrac{1}{2}. What does his number say about the circle? Find the slopes of the circle where x = 2.
  2. Moirrey differentiates x^2 + 4y^2 + 9 = 0 and gets \dfrac{dy}{dx} = -\dfrac{x}{4y}. Explain why this formula describes no slope at all.
  3. Breesha differentiates y^3 = x^2. Show that her slope formula is \dfrac{0}{0} at the origin, and use y = x^{2/3} to explain why the curve has no tangent slope there.
  4. Prove that the tangent line to the circle x^2 + y^2 = r^2 at any point (a, b) of the circle is ax + by = r^2, including the points with b = 0.
  5. Suppose y is a differentiable function of x with y^5 + y = x. Show that \dfrac{dy}{dx} > 0 at every point, find its value at (2, 1), and say what this means for the graph.
  6. Show that y^2 = x^2 defines y implicitly as a function of x near (1, 1) and near (1, -1) but not near (0, 0), and check that the implicit formula gives the right slope at the first two points.
Answers
  1. \dfrac{dy}{dx} = -\dfrac{x}{3y}, and -\dfrac{1}{2} at (3, 2). From 2x + 6y\,y' = 0; the point is on the curve since 9 + 12 = 21.
  1. \dfrac{dy}{dx} = \dfrac{2x}{3y^2 + 1}. From 3y^2\,y' + y' = 2x; the factor 3y^2 + 1 is never 0.
  1. \dfrac{dy}{dx} = \dfrac{3x^2 + y^2}{2 - 2xy}. The product rule on xy^2 gives y^2 + 2xy\,y', so 3x^2 + y^2 + 2xy\,y' = 2y'; collect the y' terms on the right.
  1. \dfrac{dy}{dx} = \dfrac{2x - y}{\cos y + x}. From \cos y\,y' + y + x\,y' = 2x.
  1. \dfrac{dy}{dx} = \dfrac{8}{3(x + y)^2} - 1. The chain rule gives 3(x + y)^2(1 + y') = 8; divide and subtract 1.
  1. y = -\dfrac{4}{5}x + \dfrac{14}{5}. From 2x + y + x\,y' + 2y\,y' = 0, y' = -\dfrac{2x + y}{x + 2y}, which is -\dfrac{4}{5} at (1, 2).
  1. y = \dfrac{2}{27}x + \dfrac{44}{27}. The point fits: 16 - 10 = 6. From 4y^3\,y' - y - x\,y' = 0, y' = \dfrac{y}{4y^3 - x} = \dfrac{2}{32 - 5}.
  1. y = -x + 2. Differentiating, 2(x^2 + y^2)(2x + 2y\,y') = 4y + 4x\,y'; at (1, 1) this is 8 + 8y' = 4 + 4y', so y' = -1. Swapping x and y leaves the equation unchanged, so the curve is its own mirror image across y = x. At (1, 1), on the mirror line, the tangent must be its own mirror image, so it is perpendicular to y = x.
  1. Tangent y = -4x + 9, normal y = \dfrac{1}{4}x + \dfrac{1}{2}. From 3x^2 + 3y^2\,y' = 0, y' = -\dfrac{x^2}{y^2} = -4; the normal has slope \dfrac{1}{4}.
  1. Horizontal at (2, 4) and (-2, -4); vertical at (4, 2) and (-4, -2). Here y' = \dfrac{y - 2x}{2y - x}. Setting y = 2x in the equation gives 3x^2 = 12, and there 2y - x = 3x \ne 0. Setting x = 2y gives 3y^2 = 12, and there y - 2x = -3y \ne 0.
  1. Horizontal at \displaystyle \left(-\dfrac{2}{\sqrt{3}}, \pm\dfrac{4}{3}\sqrt[4]{3}\right) \approx (-1.155, \pm 1.755); vertical at (-2, 0), (0, 0) and (2, 0). Here y' = \dfrac{3x^2 - 4}{2y}, whose top is 0 at \displaystyle x = \pm\dfrac{2}{\sqrt{3}}. At \displaystyle x = -\dfrac{2}{\sqrt{3}}, \displaystyle y^2 = x^3 - 4x = \dfrac{16\sqrt{3}}{9}; at \displaystyle x = \dfrac{2}{\sqrt{3}}, \displaystyle x^3 - 4x = -\dfrac{16\sqrt{3}}{9} < 0, so no real y exists. The bottom is 0 when y = 0, so x = 0, \pm 2, where the top is -4, 8, 8.
  1. (4, 2) and (-4, -2). Setting y' = -\dfrac{x}{4y} equal to -\dfrac{1}{2} gives x = 2y, so 8y^2 = 32.
  1. -36. With y' = -\dfrac{x^2}{y^2}, the quotient rule gives y'' = -\dfrac{2xy^2 - 2x^2 y\,y'}{y^4} = -\dfrac{2xy + 2x^4/y^2}{y^3}, which is -\dfrac{2x(y^3 + x^3)}{y^5}; the equation gives x^3 + y^3 = 9, so y'' = -\dfrac{18x}{y^5}, and at (2, 1) this is -36.
  1. \dfrac{d^2y}{dx^2} = -\dfrac{36}{y^3}, which is -\dfrac{1}{6} at (3, 6). From 2y\,y' = 12, y' = \dfrac{6}{y}, and y'' = -\dfrac{6y'}{y^2} = -\dfrac{36}{y^3}.
  1. \dfrac{8}{125}. From y + x\,y' + 2y\,y' = 0, y' = -\dfrac{y}{x + 2y}. The quotient rule gives y'' = \dfrac{y - x\,y'}{(x + 2y)^2}, and the top becomes \dfrac{2xy + 2y^2}{x + 2y} = \dfrac{8}{x + 2y} by the equation. At (3, 1), x + 2y = 5.
  1. f'(32) = \dfrac{1}{20} and g'(4) = -\dfrac{1}{16}. f'(x) = \dfrac{2}{5}x^{-3/5} and 32^{-3/5} = \dfrac{1}{8}; g(x) = x^{-1/2}, so g'(x) = -\dfrac{1}{2}x^{-3/2} and 4^{-3/2} = \dfrac{1}{8}.
  1. y' = \dfrac{2x}{3y^2}, and -\dfrac{2}{9} at x = -27. From 3y^2\,y' = 2x; at x = -27, y = 9, so y' = \dfrac{-54}{243}. The power rule gives \dfrac{2}{3}(-27)^{-1/3} = \dfrac{2}{3} \cdot \left(-\dfrac{1}{3}\right), the same.
  1. y = \dfrac{1}{12}x + \dfrac{4}{3}. y' = \dfrac{1}{3}x^{-2/3}, which is \dfrac{1}{12} at x = 8, with y = 2. At x = 0 the derivative does not exist: the difference quotient h^{-2/3} grows without bound, and the graph has a vertical tangent.
  1. Nothing: (2, 4) is not on the circle, since 4 + 16 = 20 \ne 25. The circle’s points with x = 2 are \displaystyle \big(2, \pm\sqrt{21}\big), with slopes \displaystyle \mp\dfrac{2}{\sqrt{21}}.
  1. x^2 + 4y^2 + 9 \ge 9 for all real x and y, so the graph is empty. The formula assumed a point of the curve, and there is none.
  1. From 3y^2\,y' = 2x, y' = \dfrac{2x}{3y^2}, which is \dfrac{0}{0} at the origin. The curve is the graph of y = x^{2/3}, whose difference quotient at 0, h^{-1/3}, grows without bound as h \to 0^+ and falls without bound as h \to 0^-: a cusp, with no tangent slope.
  1. Differentiating, y' = -\dfrac{x}{y}, so for b \ne 0 the tangent is y - b = -\dfrac{a}{b}(x - a). Multiplying by b, by - b^2 = -ax + a^2, so ax + by = a^2 + b^2 = r^2. When b = 0, a = \pm r and the tangent is vertical, x = \pm r; the formula gives \pm r\,x = r^2, which is the same line.
  1. \dfrac{dy}{dx} = \dfrac{1}{5y^4 + 1}, which is \dfrac{1}{6} at (2, 1). From 5y^4\,y' + y' = 1; the bottom is at least 1, so the slope is positive at every point and the graph rises everywhere.
  1. y^2 = x^2 is the pair of lines y = x and y = -x. Near (1, 1) the curve is just y = x, near (1, -1) just y = -x; near (0, 0) both lines are present, and no function of x has both as its graph. From 2y\,y' = 2x, y' = \dfrac{x}{y}, which is 1 and -1 at the two points, the slopes of the lines.
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