Factoring when the exponents are negative or fractional

The tidy-up step every derivative of a root demands, and the one rule that makes it work.

Expressions like $x^{1/2}+x^{-1/2}$ look unfactorable, because there is no whole power of $x$ sitting in both terms. There is, and the rule for finding it is short, but it is worth confirming the answer first so you know what is being aimed at.

So $x^{1/2}+x^{-1/2}$ really is $\frac{x+1}{\sqrt{x}}$. Here is where that comes from.

The rule: pull out the smallest exponent

The common factor of a set of powers of $x$ is $x$ raised to the smallest exponent present, where smallest means furthest left on the number line. Between $\frac{1}{2}$ and $-\frac{1}{2}$, the smaller is $-\frac{1}{2}$.

The reason is the division law. Taking out $x^{-1/2}$ means dividing each term by it, and dividing subtracts exponents. Every other exponent minus the smallest one is zero or more, so nothing is left with a negative power inside the bracket.

1. The terms are $x^{1/2}$ and $x^{-1/2}$. The exponents are $\frac{1}{2}$ and $-\frac{1}{2}$, and the smaller is $-\frac{1}{2}$.
2. Take out $x^{-1/2}$. The first term contributes $x^{1/2} \div x^{-1/2} = x^{1/2-(-1/2)} = x^{1}$.
3. The second term contributes $x^{-1/2} \div x^{-1/2} = x^{0} = 1$. A term that is entirely the common factor leaves a 1 behind, never nothing.
4. So $x^{1/2}+x^{-1/2} = x^{-1/2}\left(x+1\right)$.
5. Finally write the negative power as a fraction: $x^{-1/2} = \frac{1}{\sqrt{x}}$, so the answer is $\dfrac{x+1}{\sqrt{x}}$.
6. Check at $x = 4$: the original is $2 + \frac{1}{2} = 2.5$, and the answer is $\frac{5}{2} = 2.5$.

The version calculus actually produces

Differentiating a product involving $\sqrt{x+3}$ produces something with the shape below, and an examiner expects it as a single fraction. The common factor is a power of a whole bracket rather than of $x$, which changes nothing: the bracket plays the part $x$ was playing.

Simplify $2x(x+3)^{-1/2} + (x+3)^{1/2}$.

1. The two powers of $(x+3)$ are $-\frac{1}{2}$ and $\frac{1}{2}$, so the common factor is $(x+3)^{-1/2}$.
2. First term: $2x(x+3)^{-1/2} \div (x+3)^{-1/2} = 2x$.
3. Second term: $(x+3)^{1/2} \div (x+3)^{-1/2} = (x+3)^{1/2-(-1/2)} = (x+3)^{1} = x+3$.
4. So the expression is $(x+3)^{-1/2}\left(2x + (x+3)\right)$.
5. Collect inside the bracket: $2x + x + 3 = 3x+3$, which is $3(x+1)$.
6. Write the negative power underneath: $\dfrac{3(x+1)}{\sqrt{x+3}}$.
7. Check at $x = 1$: the original is $2 \cdot 1 \cdot \frac{1}{2} + 2 = 3$, and the answer is $\frac{3 \cdot 2}{2} = 3$.

Numbers in front, and thirds

The numerical coefficients have a common factor of their own, and it comes out at the same time. In $3x^{-2/3}+6x^{1/3}$ the numbers 3 and 6 share a 3, and the exponents $-\frac{2}{3}$ and $\frac{1}{3}$ have the smaller $-\frac{2}{3}$.

So the common factor is $3x^{-2/3}$. The first term leaves 1. The second leaves $\frac{6}{3}x^{1/3-(-2/3)} = 2x^{1} = 2x$. The result is $3x^{-2/3}(1+2x)$, or $\dfrac{3(1+2x)}{\sqrt[3]{x^{2}}}$.

Notice the shape of every answer in this lesson: a single fraction, with a root on the bottom and a tidy polynomial on the top. That is the form an examiner marks and the form a later step can use, because a single fraction can be compared with zero, cancelled against something, or read for the values that make it undefined. A sum of terms with scattered negative exponents can do none of those things.

The reason this shape keeps appearing is that the rules for differentiating a product and a root each contribute one term, and the two terms carry powers of the same bracket that differ by exactly 1. Two exponents that differ by 1 always have a common factor, so the collapse in this lesson is always available.