The slope of a curve
Tonight's idea
A straight line has one slope, the same everywhere. A curve does not: \(y = x^2\) is flat at the bottom and steep out at the sides. So what should "the slope of \(y = x^2\) at \(x = 1\)" even mean?
Here is the answer. Pick the point \((1, 1)\) and a second point a little further along, at \(x = 1 + h\). The line through two points on a curve is a secant, with slope \(\dfrac{(1 + h)^2 - 1}{h}\). Now shrink \(h\):
| \(h\) | 1 | 0.1 | 0.01 | 0.001 |
|---|---|---|---|---|
| secant slope | 3 | 2.1 | 2.01 | 2.001 |
The slopes close in on 2. The line through \((1,1)\) with slope 2 just touches the curve: the tangent. The slope of the curve at \(x = 1\) is the limit of the secant slopes as \(h \to 0\). Algebra does every point at once:
\[ \frac{(x + h)^2 - x^2}{h} = \frac{2xh + h^2}{h} = 2x + h \;\to\; 2x. \]
That formula, \(2x\), is the derivative of \(x^2\), written \(f'(x)\) or \(\dfrac{dy}{dx}\): a new function giving the slope of the old one wherever you ask. Notice the move: we could not put \(h = 0\) straight in (that gives \(0/0\)), so we simplified first and then let \(h \to 0\). That one idea is the whole subject.
Watch one: the derivative of \(x^3\), and a tangent line
\[ \frac{(x+h)^3 - x^3}{h} = \frac{3x^2h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2 \;\to\; 3x^2. \]
So the derivative of \(x^3\) is \(3x^2\). At \(x = 2\) the slope is 12, and the tangent line through \((2, 8)\) with slope 12 is \(y - 8 = 12(x - 2)\), that is \(y = 12x - 16\).
Your turn
- Find the derivative of \(f(x) = x^2 - 4x\) from first principles. Where is it zero, and what is happening on the graph there?
- Write the equations of the tangent lines to \(y = x^2\) at \(x = 3\) and at \(x = -1\). Find where the two tangents meet.
- Find the derivative of \(1/x\): simplify \(\dfrac{\frac{1}{x+h} - \frac1x}{h}\) (put the top over a common denominator first), then let \(h \to 0\). What is the slope at \(x = 2\)?
- Find the derivative of \(\sqrt x\): write \(\dfrac{\sqrt{x+h} - \sqrt x}{h}\), multiply top and bottom by \(\sqrt{x+h} + \sqrt x\), simplify, and let \(h \to 0\). What is the slope at \(x = 4\)?
Drag \(h\) toward zero and watch the secant become the tangent at mathdetours.com/demos/tangent.html.
Stretch: the pattern
Collect tonight's results: \(x^2 \to 2x\), \(x^3 \to 3x^2\), \(x^{-1} \to -x^{-2}\), \(x^{1/2} \to \tfrac12 x^{-1/2}\). State the pattern for \(x^n\) in one line. Then use it to find both points where the tangent to \(y = x^3 - 3x\) is horizontal.
Answers
- \(\dfrac{(x+h)^2 - 4(x+h) - x^2 + 4x}{h} = \dfrac{2xh + h^2 - 4h}{h} = 2x + h - 4 \to 2x - 4\). Zero at \(x = 2\), where the tangent is horizontal: the bottom of the parabola, \(f(2) = -4\).
- Slopes \(2x\): 6 and \(-2\). Through \((3, 9)\): \(y = 6x - 9\). Through \((-1, 1)\): \(y = -2x - 1\). Meet when \(6x - 9 = -2x - 1\), \(x = 1\), \(y = -3\): the point \((1, -3)\). (The meeting \(x\)-coordinate is the average of 3 and \(-1\), true for any two tangents to a parabola.)
- \(\dfrac{1}{x+h} - \dfrac1x = \dfrac{-h}{x(x+h)}\). Divide by \(h\): \(\dfrac{-1}{x(x+h)} \to -\dfrac{1}{x^2}\). At \(x = 2\): \(-\tfrac14\). Negative because \(1/x\) is falling.
- \(\dfrac{(x + h) - x}{h(\sqrt{x+h} + \sqrt x)} = \dfrac{1}{\sqrt{x+h} + \sqrt x} \to \dfrac{1}{2\sqrt x}\). At \(x = 4\): \(\tfrac14\).
Stretch. The derivative of \(x^n\) is \(nx^{n-1}\); check \(n = 2, 3, -1, \tfrac12\). For \(y = x^3 - 3x\) the derivative is \(3x^2 - 3\), zero at \(x = \pm1\): the points \((1, -2)\) and \((-1, 2)\). The first is a local low, the second a local high.
After the area sheet, try Area and slope undo each other: one thin strip connects the two ideas.