Before the midterm

How do you find a slope without solving for y?

Differentiate the equation as it stands, with y as a function of x: the slope comes out in terms of the point, fails where the curve turns vertical, and gives an inverse function’s slope as one over a reflected slope.

Builds on the chain rule and logarithms.

Point, (x, y): (1.00, 1.00)
Slope, dy/dx = −(2x + y)/(x + y): −1.50
Bottom of the fraction, x + y: 2.00

Step 1 of 5: Two points, one x

This is the curve $2x^2 + 2xy + y^2 = 5$ . It is no one function's graph: most vertical lines that meet it meet it twice. Still, at each point it has a tangent, drawn here, and so a slope. Drag the point around the curve.

Predict first. The curve passes through $(1, 1)$ and $(1, -3)$ , both at $x = 1$ . Can its slope be a formula in $x$ alone?

Point, around the curve1.107

1. Two points, one x
This is the curve $2x^2 + 2xy + y^2 = 5$ . It is no one function's graph: most vertical lines that meet it meet it twice. Still, at each point it has a tangent, drawn here, and so a slope. Drag the point around the curve.
Question: The curve passes through $(1, 1)$ and $(1, -3)$ , both at $x = 1$ . Can its slope be a formula in $x$ alone?
No. At $(1, 1)$ the slope is $-1.5$ ; at $(1, -3)$ , the same $x$ , it is $-0.5$ . Differentiate both sides with $y$ a function of $x$ : $y^2$ gives $2y\,y'$ and $2xy$ gives $2y + 2xy'$ . Solving for $y'$ gives a formula in the point, $x$ and $y$ both.
$4x + 2y + 2x\,y' + 2y\,y' = 0 \quad\Longrightarrow\quad y' = -\frac{2x + y}{x + y}$
The figure: The curve 2x² + 2xy + y² = 5, a tilted oval, with its tangent at (1.00, −3.00), of slope −0.50. The point (1.00, 1.00) is on the curve at the same x.
2. Where the slope fails
The slope is a fraction, and a fraction can fail. Watch the bottom of it, $x + y$ , under the figure as you move the point.
Drag the point to where the tangent is vertical.
At about $(2.24, -2.24)$ , and at $(-2.24, 2.24)$ on the other side, the bottom $x + y$ is $0$ : the formula divides by $0$ , there is no slope, and the tangent is vertical. Where the top, $2x + y$ , is $0$ instead, the tangent is flat.
The figure: The curve 2x² + 2xy + y² = 5, a tilted oval, with its tangent at (2.24, −2.24), which is vertical.
3. The slip: no y′
A common slip differentiates $y^2$ as $2y$ , as though $y$ were the variable. The rest is right: $4x + 2y + 2xy' + 2y = 0$ , so $y' = -2(x + y)/x$ . Its line is in amber, beside the tangent.
Question: Why must $y^2$ give $2y\,y'$ ?
$y$ is a function of $x$ , so $y^2$ is a composite and the chain rule brings out $y'$ . Without it the slope at $(1, 1)$ comes out $-4$ , the amber line, which cuts into the curve; with it, $-1.5$ , the tangent. Drag the point and compare them.
The figure: The curve 2x² + 2xy + y² = 5, a tilted oval, with its tangent at (1.00, 1.00), of slope −1.50. In amber, the line the slip gives there, of slope −4.00, which cuts across the curve.
4. An inverse, by reflection
The same method gives an inverse's slope. $y = \arctan x$ means $\tan y = x$ ; differentiate: $\sec^2 y \cdot y' = 1$ . In the figure, arctan is tan reflected in $y = x$ : $P$ and its mirror $Q$ swap coordinates, and their tangents swap rise and run.
Question: At $P = (1, \pi/4)$ , what is the slope of $y = \arctan x$ ?
$1/2$ . The mirror $Q = (\pi/4, 1)$ is where $\tan$ climbs at $\sec^2(\pi/4) = 2$ ; swapping rise and run gives $1/2$ . The amber point is the slip: $\tan$ 's slope read at $x = 1$ , $\sec^2 1 \approx 3.43$ , which gives $0.29$ , a line that cuts the curve.
The figure: The curve y = arctan x, and dashed, y = tan x, its reflection in the line y = x. P = (1.00, 0.79) on arctan and its mirror Q = (0.79, 1.00) on tan, where tan's slope is 2.00. The tangent at P has slope 0.50, one over 2.00. In amber, tan's point at x = 1.00, of slope 3.43, and the line at P of slope 0.29 it gives.
5. At every point
Since $\tan y = x$ , $\sec^2 y = 1 + \tan^2 y = 1 + x^2$ : the slope at $Q$ can be written with $P$ 's own $x$ .
Drag $P$ along $y = \arctan x$ to where its slope is $1/5$ .
At $x = 2$ . The mirror point is where $\tan$ climbs at $1 + 2^2 = 5$ , so the slope here is $1/5$ . At every point, $(\arctan x)' = 1/(1 + x^2)$ .
$\frac{d}{dx}\arctan x = \frac{1}{\sec^2 y} = \frac{1}{1 + x^2}$
The figure: The curve y = arctan x, and dashed, y = tan x, its reflection in the line y = x. P = (2.00, 1.11) on arctan and its mirror Q = (1.11, 2.00) on tan, where tan's slope is 5.00. The tangent at P has slope 0.20, one over 5.00.

Step 1 of 5: Two points, one x

The lesson as text

1. Two points, one x

2. Where the slope fails

3. The slip: no y′

4. An inverse, by reflection

5. At every point