Rewrite powers before differentiating

One function, two forms

The chalk curve is the fraction \(\dfrac{x^3 + 6x}{x^2}\). The power rule does not apply to a fraction, so first divide each term on the top by \(x^2\). Subtracting exponents gives \(\dfrac{x^3}{x^2} + \dfrac{6x}{x^2} = x + 6x^{-1}\).

Select Rewrite the fraction. The orange dashed curve of \(x + 6x^{-1}\) lies on the chalk curve, so the two forms are the same function for \(x > 0\).

The rewritten form is a sum of powers, so the power rule gives its slope, \(1 - 6x^{-2}\). Move the orange point with the slider. At \(x = 2\), both forms have the value 5, and the slope of the tangent is \(1 - 6 \cdot 2^{-2} = -\tfrac12\). The curves are identical, so this is also the slope of the fraction.

Any fraction with a single term below

The same works for any fraction with a sum on the top and a single term below, such as \(x^2\) or \(\sqrt{x}\). Select Show another fraction. The rewritten curve again lies on the original curve.

A power of \(x\) below becomes a negative exponent, and a root becomes a fractional exponent. The rewritten form is a sum of constants times powers of \(x\), so the power rule applies to each term.

The two curves are identical, so they have the same derivative. Choose the form that is easier to differentiate.

Notation

1. Divide each term of the numerator by the denominator. This splits one fraction into several terms.

   \[\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}\]

   The top splits, but a sum on the bottom does not: with \(a = b = 1\), \(\tfrac{1}{a+b} = \tfrac12\), but \(\tfrac1a + \tfrac1b = 2\). This means the method needs a denominator with a single term.

2. The laws of exponents turn each term into a constant times a power of \(x\).

   \[\begin{aligned} \frac{x^a}{x^b} &= x^{a-b} \\ \frac{1}{x^b} &= x^{-b} \\ \sqrt[m]{x} &= x^{1/m} \end{aligned}\]

3. Differentiate term by term with the sum, constant multiple and power rules.

   \[\frac{d}{dx}\bigl[\,x + 6x^{-1}\,\bigr] = 1 - 6x^{-2}\]

4. Write the answer in the form that the question uses. A negative exponent means a fraction.

   \[1 - 6x^{-2} = 1 - \frac{6}{x^2}\]

When it fails

The rewrite must keep the domain. The fraction \(\dfrac{x^3 + 2x}{x}\) has no value at \(x = 0\), because the bottom is 0, but \(x^2 + 2\) has the value 2 there.

The curves match everywhere except at the open circle \((0, 2)\). The original function has no derivative at \(x = 0\), so the slope \(2x\) holds only for \(x \ne 0\).

Example 1: a root below

Find the derivative of \(y = \dfrac{x^2 + 3}{\sqrt{x}}\) for \(x > 0\).

1. Write the root as a power, and divide each term on the top by \(x^{1/2}\).

   \[y = \frac{x^2}{x^{1/2}} + \frac{3}{x^{1/2}}\]

2. Subtract exponents in the first term: \(2 - \tfrac12 = \tfrac32\). The second term has \(x^{1/2}\) below, so its exponent is \(-\tfrac12\).

   \[y = x^{3/2} + 3x^{-1/2}\]

3. Differentiate each term. In the second term, the exponent \(-\tfrac12\) multiplies the 3 in front, which gives \(-\tfrac32\).

   \[y' = \tfrac32\,x^{1/2} - \tfrac32\,x^{-3/2}\]

4. At \(x = 1\), the slope is \(\tfrac32 - \tfrac32 = 0\). The tangent is flat there, at the lowest point of the curve.

Example 2: a coefficient below

Find the derivative of \(y = \dfrac{x^3 + 8}{2x}\).

1. Divide each term on the top by the denominator \(2x\).

2. A tempting wrong move: Move the 2 to the top: \(\dfrac{x^3}{2x} = 2x^2\).

   \[y = 2x^2 + 4x^{-1}\]

   At \(x = 2\), the fraction is \(\dfrac{8 + 8}{4} = 4\). The red curve \(2x^2 + 4x^{-1}\) is \(8 + 2 = 10\) there, far above the chalk curve, so it is a different function.

3. A 2 in the denominator divides, so \(\dfrac{x^3}{2x} = \tfrac12 x^2\) and \(\dfrac{8}{2x} = 4x^{-1}\).

   \[y = \tfrac12 x^2 + 4x^{-1}\]

4. Differentiate each term.

   \[y' = x - 4x^{-2}\]

5. At \(x = 2\), the slope is \(2 - 4 \cdot 2^{-2} = 1\), so the orange tangent rises.

Practice

Find the derivative of \(y = \dfrac{2x + 5}{x^3}\).

The dashed curve lies on the original curve. Write the formula of the dashed curve first: divide each term on the top by \(x^3\).

Questions

The graph shows \(y = \dfrac{x^3 + 16}{x}\) for \(x > 0\) and, below it, its slope. The curve is red where it falls and green where it rises. Questions 1 and 2 use the graph. Question 3 needs a calculation.

1. At which \(x\) is the tangent horizontal?

   • \(x = 2\)
   • \(x = 12\)
   • \(x = 16\)
   Answer

   \(x = 2\)

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

   • Positive
   • Negative
   • Zero
   Answer

   Negative

3. Find the derivative of \(y = \dfrac{5}{3\sqrt{x}}\) for \(x > 0\).

   • \(-\tfrac56\,x^{-3/2}\)
   • \(-\tfrac{15}{2}\,x^{-3/2}\)
   • \(-\tfrac56\,x^{-1/2}\)
   • \(\tfrac56\,x^{-1/2}\)
   Answer

   \(-\tfrac56\,x^{-3/2}\)

Summary

You rewrite a fraction or a root as constants times powers of \(x\), and then differentiate term by term.

This works because the rewritten form is the same function, so it has the same tangent at every point.

You use it when the denominator is a single term, such as \(x^2\) or \(\sqrt{x}\).

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