Use the derivative at a point

The slope changes along a curve

The curve is \(y = x^3 - 12x\), green where it rises and red where it falls. The orange line is the tangent at the orange point. The lower graph is the slope curve, so the height of its orange bar is the slope at the point.

Go to \(x = 0\). The curve is red, and the slope is \(-12\). Go to \(x = 3\). The curve is green, and the slope is 15.

Go to \(x = 2\). The tangent is horizontal, and the slope is 0. The same happens at \(x = -2\).

The slope curve is the derivative \(3x^2 - 12\). Substitute any \(x\) in it to find the slope at that point.

Three questions, one derivative

For a curve made of power terms, the derivative \(f'\) gives the slope at every point.

Three questions use it. Given \(x = a\), substitute \(a\) in \(f'\) to find the slope. Given a slope \(m\), solve \(f'(x) = m\) to find the points with that slope. Given a point and a slope, write the tangent line.

The dashed line on the curve has slope \(m\). Drag the point until the tangent lies on it. At that \(x\), the slope curve meets the dashed horizontal line at height \(m\), because its height is the slope.

Notation

1. The slope of the curve at \(x = a\) is \(f'(a)\). Differentiate each term. Then substitute \(a\).

   \[\cy{f'(a)} = \text{slope at } x = a\]

2. The tangent line touches the curve at the point \((a, f(a))\) and has the slope \(f'(a)\). Its equation in point-slope form is:

   \[y = f(a) + \cy{f'(a)}\,(x - a)\]

3. To find where the slope equals a number \(m\), solve this equation for \(x\). Then find \(y\) from \(f\). For \(m = 15\), \(3x^2 - 12 = 15\) gives \(x = 3\) or \(x = -3\).

   \[f'(x) = m\]

4. If a coefficient such as \(k\) is unknown, keep it in \(f'\). The slope condition then becomes an equation for \(k\).

   \[\begin{aligned} f(x) &= x^2 + kx \\ f'(2) &= 4 + k \end{aligned}\]

When it fails

The tangent line stays close to the curve only near \(a\), so the red gap between them grows as the second point moves away.

This gap is the error if you use the line in place of the curve. With \(a = 1\), it is about 0.03 at \(x = 1.1\) but 20 at \(x = 3\).

Example 1: a slope and a tangent line

Find the slope of \(y = x^4 - 6x\) at \(x = 1\). Then write the equation of the tangent line there.

1. Differentiate each term with the power rule.

   \[y' = 4x^3 - 6\]

2. Substitute \(x = 1\) to find the slope.

   \[y'(1) = 4 - 6 = \cy{-2}\]

3. Substitute \(x = 1\) in \(y\) to find the height of the point.

   \[y(1) = 1 - 6 = -5\]

4. The tangent line passes through \((1, -5)\) with slope \(-2\), so it falls, like the red curve at \(x = 1\).

   \[y = -5 - 2(x - 1) = -2x - 3\]

Example 2: where the slope is zero

Find the points where the curve \(y = x^3 - 3x^2 + 2\) has a horizontal tangent.

1. A horizontal tangent has slope 0, so set the derivative equal to 0.

   \[y' = 3x^2 - 6x = 0\]

2. A tempting wrong move: Divide both sides by \(x\).

   \[3x = 6 \;\Rightarrow\; x = 2\]

   The curve is flat in two places, a peak at \(x = 0\) and a valley at \(x = 2\). Dividing by \(x\) divides by 0 when \(x = 0\), so it loses the peak.

3. Factor instead. A product is 0 when one of its factors is 0.

   \[3x(x - 2) = 0 \;\Rightarrow\; x = 0 \text{ or } x = 2\]

4. Substitute each \(x\) in \(y\) to find the height.

   \[(0,\, 2) \text{ and } (2,\, -2)\]

Practice

The curve \(y = x^3 + kx\) has slope 7 at \(x = 1\). Find \(k\).

Slide \(k\) until the orange tangent at \(x = 1\) lies on the dashed line. Then write the slope at \(x = 1\) with \(k\) left in.

Questions

The graph shows \(y = x^4 - 2x^2\) and its slope. Questions 1 and 2 use it. Question 3 needs a calculation.

1. How many points of the curve have a horizontal tangent?

   • 1
   • 2
   • 3
   • 4
   Answer

   3

2. What is the slope of the curve at \(x = -2\)?

   • 24
   • −24
   • 8
   Answer

   −24

3. The curve \(y = cx^2 - x^3\) has slope 4 at \(x = 2\). Find \(c\).

   • 4
   • 1
   • −2
   • 16
   Answer

   4

Summary

You find the slope at a point by differentiating each term and substituting the input.

This works because the slope at a point is the derivative at that x, so a slope condition becomes an equation to solve.

You use the tangent line only near its point of contact, because the gap between the curve and the line grows as you move away.

This is one of four skills in topic 2.6 of Which & Why: AP Calculus.