Chapter 5 · Continuity and Differentiability

Exercise 5.3 · Q5: $\frac{dy}{dx}$ if $x^2+xy+y^2=100$

Exercise 5.3Implicit differentiation · Section 5.3.2

The question

Find $\dfrac{dy}{dx}$ in the following:

x^2 + xy + y^2 = 100

Exam answer

Differentiating both sides w.r.t. $x$ ⁠,

\begin{aligned} &2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} \\ &= 0 \end{aligned}

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

$\therefore\ \dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}$ $(x + 2y \ne 0)$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The relation describes a closed curve (a tilted ellipse), so $y$ is not one function of $x$ on the whole curve; but near most of its points the curve is the graph of a differentiable function, and implicit differentiation gives its slope there. Differentiate both sides with respect to $x$ ⁠, using the product rule on $xy$ and the chain rule on $y^2$ ⁠; the derivative of the constant $100$ is $0$ ⁠. The formula fails exactly where the curve has a vertical tangent.

Given, to find, and the plan

Given: $x^2 + xy + y^2 = 100$ ⁠.
To find: $\dfrac{dy}{dx}$ ⁠.
Plan: Differentiate term by term, collect the $\frac{dy}{dx}$ terms, factor, divide; then find the points where the divisor vanishes.

Basics you need

1Differentiating a relation

Both sides of a relation that holds along a piece of curve have equal derivatives there; the derivative of a constant is $0$ (Example 22 uses exactly this). Lesson: implicit differentiation

2Product rule

$\frac{d}{dx}(xy) = 1\cdot y + x\,\frac{dy}{dx}$ (rule (2), Section 5.3). Basics: derivative rules

3Chain rule on $y^2$

$\frac{d}{dx}(y^2) = 2y\,\frac{dy}{dx}$ ⁠. Lesson: the chain rule

4Collecting and factoring

Gather the $\frac{dy}{dx}$ terms, factor, divide by the bracket where it is non-zero. Basics: solving for dy/dx

Solution

Differentiate each term with respect to $x$
$\begin{aligned} &\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) \\ &\quad = \frac{d}{dx}(100). \end{aligned}$
- $\frac{d}{dx}(x^2) = 2x$ ⁠.
- $\frac{d}{dx}(xy) = y + x\,\frac{dy}{dx}$ (product rule).
- $\frac{d}{dx}(y^2) = 2y\,\frac{dy}{dx}$ (chain rule).
- $\frac{d}{dx}(100) = 0$ ⁠.
So
$2x + y + x\,\frac{dy}{dx} + 2y\,\frac{dy}{dx} = 0.$
Along the curve the left side is always $100$ ⁠, a constant, so its rate of change is $0$ ⁠. The middle term needs the product rule because both $x$ and $y$ change.

Product rule (2), Section 5.3; chain rule, Theorem 4; derivative of a constant is $0$ ⁠. Basics: derivative rules
Collect and factor
$\begin{aligned} x\,\frac{dy}{dx} + 2y\,\frac{dy}{dx} &= -2x - y, \\ (x + 2y)\,\frac{dy}{dx} &= -(2x + y). \end{aligned}$
Keep the two $\frac{dy}{dx}$ terms on the left, move $2x + y$ to the right (changing its sign), and factor out $\frac{dy}{dx}$ ⁠.

Factoring the common unknown. Basics: solving for dy/dx
Divide, where allowed

Wherever $x + 2y \ne 0$ ⁠,
$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}.$
The bracket $x + 2y$ is the divisor. The next step finds the two points of the curve where it is zero.
Where is $x + 2y = 0$ on the curve?

Put $x = -2y$ into the relation:
$\begin{aligned} 4y^2 - 2y^2 + y^2 &= 100, \\ 3y^2 &= 100, \\ y &= \pm\frac{10}{\sqrt3}. \end{aligned}$
So the points are $A\left(\frac{20}{\sqrt3}, -\frac{10}{\sqrt3}\right)$ and $B\left(-\frac{20}{\sqrt3}, \frac{10}{\sqrt3}\right)$ ⁠, approximately $(11.55, -5.77)$ and $(-11.55, 5.77)$ ⁠. There Step 2 reads $0\cdot\frac{dy}{dx} = -(2x + y) = \mp\frac{30}{\sqrt3} \ne 0$ ⁠, which no number satisfies: the tangent is vertical, and $\frac{dy}{dx}$ does not exist.

These are the rightmost and leftmost points of the ellipse (see the figure). There the curve turns back, so near $A$ or $B$ it is not the graph of a single function of $x$ ⁠. Everywhere else on the curve the formula holds.

Substituting one equation into another to find common points (Class X). Basics: solving for dy/dx

The curve

x^2 + xy + y^2 = 100

is an ellipse tilted at

45^\circ

(same scale on both axes). At

P(10, 0)

the tangent has slope

-\frac{2\cdot10 + 0}{10 + 0} = -2

(gold). The dashed line

x + 2y = 0

meets the ellipse at

A

and

B

⁠, where the tangents are vertical and

\frac{dy}{dx}

does not exist.

Result

\frac{dy}{dx} = -\frac{2x + y}{x + 2y}, \quad x + 2y \ne 0.

(Textbook answer: $-\dfrac{(2x + y)}{(x + 2y)}$ ⁠, the same.)

Check it with numbers

At $x = 10$ the relation gives $100 + 10y + y^2 = 100$ ⁠, so $y = 0$ or $y = -10$ ⁠. The formula gives slope $-\frac{20}{10} = -2$ at $(10, 0)$ and $-\frac{20 - 10}{10 - 20} = 1$ at $(10, -10)$ ⁠; at $(0, 10)$ it gives $-\frac{10}{20} = -\frac12$ ⁠.

Directly: solving the quadratic in $y$ ⁠, the upper and lower halves of the ellipse are $y = \frac{-x \pm \sqrt{400 - 3x^2}}{2}$ ⁠. Difference quotients with $h = 0.0001$ give $-2.0000000$ at $x = 10$ on the upper half (through $(10, 0)$ ⁠), $1.0000000$ on the lower half (through $(10, -10)$ ⁠), and $-0.5000000$ at $x = 0$ on the upper half. All agree.

Common mistakes

$\frac{d}{dx}(xy) = x\frac{dy}{dx}$ ⁠. The term $y$ from the product rule is missing; it is what turns $2x$ into $2x + y$ ⁠.
$\frac{d}{dx}(100) = 100$ ⁠. The derivative of a constant is $0$ ⁠.
Losing the minus sign. The answer is $-\frac{2x + y}{x + 2y}$ ⁠; check it at $(10, 0)$ ⁠, where the curve is falling: just to the left, at $x = 9.9$ ⁠, the upper half has $y \approx 0.20 > 0$ ⁠.

What to take away

Implicit differentiation gives the slope at every point of a curve, even when the curve is not the graph of one function.
The points where the denominator of $\frac{dy}{dx}$ vanishes are worth finding: on a closed curve they are the points with vertical tangents.

Every fact used, and where it comes from

Fact	Used in	Source
Differentiating both sides of a relation; derivative of a constant is $0$	Step 1	Section 5.3.2, Example 22 (Lesson: implicit differentiation)
Product rule $(uv)' = u'v + uv'$	Step 1	Rule (2), Section 5.3
Chain rule; $(x^n)' = nx^{n-1}$	Step 1	Theorem 4, Section 5.3.1; Table 5.3
Quadratic formula	Check	Class X, Quadratic Equations

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Exercise 5.3 · Q5: dydx\frac{dy}{dx}dxdy​ if x2+xy+y2=100x^2+xy+y^2=100x2+xy+y2=100

Basics you need

Solution

Differentiate each term with respect to xxx

Collect and factor

Divide, where allowed

Where is x+2y=0x + 2y = 0x+2y=0 on the curve?

Every fact used, and where it comes from

Exercise 5.3 · Q5: $\frac{dy}{dx}$ if $x^2+xy+y^2=100$

Differentiate each term with respect to $x$

Where is $x + 2y = 0$ on the curve?