Differentiate negative and fractional powers

Slopes of 1/x² and √x

The top graph shows \(y = 1/x^2\) for \(x > 0\), with an orange tangent at the point. The curve is red because it falls. The orange bar on the lower graph is as tall as the slope, so it points down.

At \(x = 1\) the slope is \(-2\), and at \(x = 2\) it is only \(-\tfrac14\), because the curve becomes flatter as \(x\) grows.

Write \(1/x^2\) as \(x^{-2}\). The power rule brings the exponent \(-2\) down in front, and the new exponent is \(-3\). At \(x = 1\) this gives \(-2 \cdot 1^{-3} = -2\), and at \(x = 2\) it gives \(-2 \cdot 2^{-3} = -\tfrac14\), the same slopes as on the graph.

Show \(\sqrt{x}\), which is \(x^{1/2}\). The rule gives \(\tfrac12 x^{-1/2}\). At \(x = 4\) the power \(4^{-1/2}\) is \(\tfrac12\), so the slope is \(\tfrac12 \cdot \tfrac12 = \tfrac14\), the height of the orange bar.

Any constant exponent

Slide the exponent \(n\) from \(-3\) to 3. The graph shows only \(x > 0\), because a power such as \(x^{1/2}\) is not defined for negative \(x\).

If \(n\) is negative, the curve is red and falls, and the orange bar points down. If \(n\) is positive, the curve is green and rises, and the bar points up.

The two numbers below the graph are the slope and \(n\) times \(x^{n-1}\). They are equal for every \(n\) and every \(x\), so the power rule works for any constant exponent.

Notation

1. The power rule needs the form \(x^n\), so first write a reciprocal or a root as a power with the laws of exponents.

   \[\frac{1}{x^m} = x^{-m}, \qquad \sqrt[m]{x} = x^{1/m}\]

2. The power rule holds for any constant exponent \(n\), not only for positive whole numbers.

   \[\frac{d}{dx}\,x^n = \cy{n}\,x^{\cy{n-1}}\]

   The proof for other exponents uses rules that come later. On the AP exam, you use the rule as stated.

3. Subtract 1 from the exponent. A negative exponent becomes more negative, and a fraction changes by a whole 1.

   \[-3 - 1 = -4, \qquad \tfrac12 - 1 = -\tfrac12\]

4. A negative exponent means a fraction, so you can write the answer as a fraction or a root again. The derivative exists only where the new power is defined, which for \(\sqrt{x}\) means \(x > 0\).

   \[\begin{aligned} -3x^{-4} &= -\frac{3}{x^4} \\ \tfrac12 x^{-1/2} &= \frac{1}{2\sqrt{x}} \end{aligned}\]

When it fails

The slope of \(\sqrt{x}\) is \(\dfrac{1}{2\sqrt{x}}\). At \(x = 0\) this divides by 0, so the formula has no value.

Move the point toward 0. The tangent becomes steeper, and at 0 it is vertical. A vertical line has no slope, so the derivative does not exist at \(x = 0\).

Example 1: a reciprocal

Find the derivative of \(y = \dfrac{1}{x^3}\).

1. Write the reciprocal as a power of \(x\), because the rule needs that form.

   \[y = x^{-3}\]

2. The exponent \(-3\) comes down in front, and the new exponent is \(-3 - 1 = -4\).

   \[y' = \cy{-3}\,x^{\cy{-4}}\]

3. Write the answer as a fraction again. The lower graph is the graph of this slope.

   \[y' = -\frac{3}{x^4}\]

4. The curve falls on both sides of 0. Because \(x^4\) is positive for every \(x \ne 0\), the slope \(-3/x^4\) is negative there. At \(x = 0\), neither the curve nor the slope exists.

Example 2: a negative fraction exponent

Find the derivative of \(y = \dfrac{1}{\sqrt{x}}\) for \(x > 0\).

1. Write the root and the reciprocal as one power.

   \[y = x^{-1/2}\]

2. A tempting wrong move: Subtract 1 from the exponent: \(-\tfrac12 - 1 = +\tfrac12\).

   \[y' = -\tfrac12\,x^{1/2}\]

   At \(x = 0.25\) this formula gives \(-\tfrac12 \cdot \tfrac12 = -\tfrac14\), the red dashed line, which is almost flat. But the curve is steep there.

3. Subtracting 1 from a negative number makes it more negative: \(-\tfrac12 - 1 = -\tfrac32\).

   \[y' = -\tfrac12\,x^{-3/2}\]

4. At \(x = 0.25 = \tfrac14\), the power \(x^{-3/2}\) is \(4^{3/2} = 8\), so the slope is \(-\tfrac12 \cdot 8 = -4\). This matches the steep orange tangent.

Practice

Find the derivative of \(y = \sqrt[4]{x}\) for \(x > 0\).

Move the point to \(x = 1\). Any power of 1 is 1, so the slope there equals the number in front of the power.

Questions

Questions 1 and 2 use the graph of \(y = 1/x\) and its slope. Question 3 needs a calculation.

1. At \(x = -2\), is the slope of the curve positive, negative or zero?

   • Positive
   • Negative
   • Zero
   Answer

   Negative

2. At \(x = 0.5\), the orange bar on the lower graph reaches \(-4\). Which formula gives the height of the lower graph?

   • \(-x^{-2}\)
   • \(-x^{-1}\)
   • \(x^{-2}\)
   • \(-2x^{-3}\)
   Answer

   \(-x^{-2}\)

3. Find the derivative of \(y = \sqrt{x^7}\) for \(x > 0\).

   • \(\tfrac72\,x^{5/2}\)
   • \(\tfrac72\,x^{7/2}\)
   • \(\tfrac72\,x^{3}\)
   • \(\tfrac12\,x^{5/2}\)
   Answer

   \(\tfrac72\,x^{5/2}\)

Summary

You differentiate a reciprocal or a root by writing it as a power of \(x\) and applying the power rule.

This works because the rule \(nx^{n-1}\) holds for every constant exponent, not only for positive whole numbers.

The derivative exists only where the new power is defined, so the slope of \(\sqrt{x}\) exists only for \(x > 0\).

This is one of four skills in topic 2.5 of Which & Why: AP Calculus.