Differentiating Roots and Fractional Powers

Rewrite √x, 1/√x, (…)/√x and 1/x^n as powers, then differentiate

In the exam for Edexcel International AS Maths · 5 real past-paper questions below

Everything as a power

You can already differentiate any power of $x$ , and you can write $\frac{125}{x}$ as $125 x^{- 1}$ . AS questions add roots and fractions with $x$ on the bottom. The rule is the same: make every term a number times a power of $x$ , then differentiate.

Formula

n x^{m} = x^{\frac{m}{n}}, \frac{1}{x ^{k}} = x^{- k}

$n$ — the root goes on the bottom of the fraction: a square root is ½, a cube root is ⅓.
$m$ — the power inside the root goes on top.
$k$ — a power on the bottom of a fraction moves up and turns negative.

You see	As a power
$x$	$x^{\frac{1}{2}}$
$\frac{1}{x}$	$x^{- \frac{1}{2}}$
$x x$	$x^{1} \times x^{\frac{1}{2}} = x^{\frac{3}{2}}$
$\frac{3}{2 x ^{2}}$	$\frac{3}{2} x^{- 2}$

Worked example

1
$y = \frac{x ^{2} - 12}{4 x}, x > 0$ . Find $\frac{d y}{d x}$ .
2
$y = \frac{x ^{2}}{4 x ^{\frac{1}{2}}} - \frac{12}{4 x ^{\frac{1}{2}}}$ — One term on the bottom, so split the fraction: divide each top term by it.
3
$y = \frac{1}{4} x^{\frac{3}{2}} - 3 x^{- \frac{1}{2}}$ — Dividing subtracts powers: $2 - \frac{1}{2} = \frac{3}{2}$ . And $12 \div 4 = 3$ , with $x^{\frac{1}{2}}$ moving up as $x^{- \frac{1}{2}}$ .
4
$\frac{1}{4} x^{\frac{3}{2}} \to \frac{3}{8} x^{\frac{1}{2}}$ — $\frac{3}{2} \times \frac{1}{4} = \frac{3}{8}$ . Take 1 off the power: $\frac{3}{2} - 1 = \frac{1}{2}$ .
5
$- 3 x^{- \frac{1}{2}} \to + \frac{3}{2} x^{- \frac{3}{2}}$ — $- \frac{1}{2} \times (- 3) = + \frac{3}{2}$ . Take 1 off: $- \frac{1}{2} - 1 = - \frac{3}{2}$ .
6
$\frac{d y}{d x} = \frac{3}{8} x^{\frac{1}{2}} + \frac{3}{2} x^{- \frac{3}{2}}$ — Put the two terms together.

Rule

Marks: the first M1 is for writing $y$ as separate terms with at least one correct power. Differentiating the top and the bottom of the fraction separately earns no method marks.

Common mistake

Writing $\frac{1}{2 x}$ as $2 x^{- \frac{1}{2}}$ . Only the $x$ moves up. The 2 stays on the bottom: $\frac{1}{2} x^{- \frac{1}{2}}$ .

Memory trick

Roots live on the bottom floor of the power: √ is ½, ∛ is ⅓. Anything that climbs up from the bottom of a fraction picks up a minus sign.

Key idea

Rewrite first, differentiate second. Every mark in the question depends on getting the powers right before you start.

Simplest form

“Find dy/dx, giving your answer in simplest form” is common. The last A1 is lost for an untidy answer, even when the calculus is right.

Worked example

1
$y = 4 x x - \frac{6}{x} + \frac{5}{2 x}, x > 0$ . Find $\frac{d y}{d x}$ in simplest form.
2
$y = 4 x^{\frac{3}{2}} - 6 x^{- \frac{1}{2}} + \frac{5}{2} x^{- 1}$ — Every term as a number times a power of $x$ .
3
$4 x^{\frac{3}{2}} \to 6 x^{\frac{1}{2}}$ — $\frac{3}{2} \times 4 = 6$ .
4
$- 6 x^{- \frac{1}{2}} \to + 3 x^{- \frac{3}{2}}$ — $- \frac{1}{2} \times (- 6) = + 3$ . Two minuses make a plus.
5
$\frac{5}{2} x^{- 1} \to - \frac{5}{2} x^{- 2}$ — $- 1 \times \frac{5}{2} = - \frac{5}{2}$ .
6
$\frac{d y}{d x} = 6 x^{\frac{1}{2}} + 3 x^{- \frac{3}{2}} - \frac{5}{2} x^{- 2}$ — Simplest form: each sign written once, each number worked out, each power a single number.
7
$\frac{d y}{d x} = 6 x + \frac{3}{x x} - \frac{5}{2 x ^{2}}$ — This is also accepted. Powers or roots, either is fine.

Steps

1
- Work out every index: $x^{\frac{1}{2}}$ , not $x^{\frac{3}{2} - 1}$ .
- One sign per term: $+ 3$ , not $- (- 3)$ or $+ - 3$ .
- Work out every number: $6$ , not $\frac{3}{2} \times 4$ .
- No $x^{1}$ or $x^{0}$ left behind: write $x$ and drop the $x^{0}$ .
- Everything on one line, as one answer.

Common mistake

Going on after a correct answer and making a slip, for example writing it over a common denominator wrongly. When a question asks for simplest form, examiners may mark your last line, so stop when it is tidy.

Key idea

Simplest form means nothing left to work out. If a number or a power could still be calculated, calculate it.

Gradient equals a number

Questions ask where the gradient is 0, or another number. With fractional powers you cannot just factorise. Two moves cover them all.

Worked example

1
$y = \frac{2}{5} x^{\frac{5}{2}} - 3 x, x > 0$ . Find the $x$ coordinate where the gradient is 24.
2
$\frac{d y}{d x} = x^{\frac{3}{2}} - 3$ — $\frac{5}{2} \times \frac{2}{5} = 1$ , and $- 3 x \to - 3$ .
3
$x^{\frac{3}{2}} - 3 = 24$ — Set the gradient equal to 24.
4
$x^{\frac{3}{2}} = 27$ — Get the power of $x$ on its own.
5
$x = 2 7^{\frac{2}{3}}$ — Undo the power $\frac{3}{2}$ with its flip, $\frac{2}{3}$ .
6
$x = (327)^{2} = 3^{2} = 9$ — Cube root first, then square. Small numbers, no calculator needed.

When there are negative powers too, multiply the whole equation by $x$ to a power that clears them. This is the “show that” version.

Worked example

1
$f (x) = \frac{x ^{2} + 5 x + 12}{x}, x > 0$ . Show that $f^{'} (x) = 0$ gives $3 x^{2} + 5 x - 12 = 0$ , and solve it.
2
$f (x) = x^{\frac{3}{2}} + 5 x^{\frac{1}{2}} + 12 x^{- \frac{1}{2}}$ — Split the fraction. Each power drops by $\frac{1}{2}$ .
3
$f^{'} (x) = \frac{3}{2} x^{\frac{1}{2}} + \frac{5}{2} x^{- \frac{1}{2}} - 6 x^{- \frac{3}{2}}$ — Differentiate each term: $- \frac{1}{2} \times 12 = - 6$ .
4
$\frac{3}{2} x^{\frac{1}{2}} + \frac{5}{2} x^{- \frac{1}{2}} - 6 x^{- \frac{3}{2}} = 0$ — Set $f^{'} (x) = 0$ . Write this line: it is where the mark comes from.
5
$\times 2 x^{\frac{3}{2}} : 3 x^{2} + 5 x - 12 = 0$ — The most negative power is $- \frac{3}{2}$ , so multiply by $x^{\frac{3}{2}}$ . The 2 clears the halves. Add powers: $\frac{1}{2} + \frac{3}{2} = 2$ , $- \frac{1}{2} + \frac{3}{2} = 1$ , $- \frac{3}{2} + \frac{3}{2} = 0$ .
6
$(3 x - 4) (x + 3) = 0$ — Factorise.
7
$x = \frac{4}{3}$ — $x = - 3$ is rejected, because the question says $x > 0$ .

Rule

“Show that” needs every line, including “ $= 0$ ” before you multiply. Writing $f^{'} (x) = 3 x^{2} + 5 x - 12$ loses the final mark: that is not $f^{'} (x)$ , it is $f^{'} (x) = 0$ multiplied by $2 x^{\frac{3}{2}}$ .

Common mistake

Giving $x = \pm 9$ or keeping $x = - 3$ . When $x > 0$ , or a power like $x^{\frac{3}{2}}$ is involved, only the positive answer counts.

Key idea

Power of $x$ on its own: raise both sides to the flipped power. Negative powers: multiply through by $x$ to the most negative power, made positive.

Hidden quadratics

After you differentiate, the equation can be a quadratic in disguise. Spot it, give it a new letter, and solve it like any quadratic.

Worked example

1
$y = x^{5} - 10 x^{3} + 7 x$ . Find the $x$ coordinates of the points where the gradient is 42.
2
$\frac{d y}{d x} = 5 x^{4} - 30 x^{2} + 7$ — Differentiate each term.
3
$5 x^{4} - 30 x^{2} - 35 = 0$ — Set it equal to 42 and take 42 from both sides.
4
$x^{4} - 6 x^{2} - 7 = 0$ — Divide every term by 5.
5
$(x^{2})^{2} - 6 x^{2} - 7 = 0$ — $x^{4} = (x^{2})^{2}$ , so this is a quadratic in $x^{2}$ .
6
$(x^{2} - 7) (x^{2} + 1) = 0$ — Factorise, treating $x^{2}$ as the unknown.
7
$x^{2} = 7 or x^{2} = - 1$ — A square cannot be negative, so $x^{2} = - 1$ gives no points.
8
$x = 7 or x = - 7$ — Here $x$ can be negative, so keep both roots.

Powers of $x$ and $x$ together hide a quadratic in $x$ . Let $u = x$ , so $x = u^{2}$ .

Worked example

1
$y = \frac{1}{2} x^{2} - 4 x x + 8 x, x > 0$ . Find the points where the gradient is 0.
2
$y = \frac{1}{2} x^{2} - 4 x^{\frac{3}{2}} + 8 x$ — Write $x x$ as a power.
3
$\frac{d y}{d x} = x - 6 x^{\frac{1}{2}} + 8$ — $\frac{3}{2} \times (- 4) = - 6$ .
4
$u^{2} - 6 u + 8 = 0$ — Let $u = x$ . Then $x = u^{2}$ and $x^{\frac{1}{2}} = u$ .
5
$(u - 2) (u - 4) = 0, u = 2 or u = 4$ — Factorise.
6
$x = 2 \Rightarrow x = 4, x = 4 \Rightarrow x = 16$ — Go back to $x$ : square each value of $u$ .
7
$(4, 8) and (16, 0)$ — Put each $x$ into $y$ : $8 - 32 + 32 = 8$ and $128 - 256 + 128 = 0$ .

Rule

Marks: one for the quadratic with 0 on one side, one for solving it with working shown. Answers just written down from a calculator lose the method mark. You must also find both $x$ values to get the final A1.

Common mistake

Stopping at $u = 2$ and $u = 4$ . Those are values of $x$ , not $x$ . Square them.

Key idea

Powers $x^{4}$ and $x^{2}$ : let the new letter be $x^{2}$ . Powers $x$ and $x^{\frac{1}{2}}$ : let it be $x$ . Then always go back to $x$ .

Which method? Read the question

Read the question and the powers you get after differentiating.

You see	You do
$x$ , $\frac{1}{x}$ or $\frac{1}{x ^{n}}$	Rewrite as powers of $x$ first
A fraction with one term on the bottom	Split it: divide each top term
“In simplest form”	Work out every index and number; one sign per term
Gradient $= k$ , one power of $x$ left	Get the power alone, raise to the flipped power
Negative powers in $f^{'} (x) = 0$	Multiply by $x$ to the most negative power, made positive
$x^{4}$ and $x^{2}$ after differentiating	Quadratic in $x^{2}$
$x$ and $x$ after differentiating	Let $u = x$ , then square back

Common mistake

Forgetting $x > 0$ at the end. Check each answer against the condition in the question.

Key idea

John Wallis worked with negative and fractional powers in 1656. In a letter of 1676, Isaac Newton wrote them the way we do today, like $x^{\frac{3}{2}}$ . With that notation, roots follow the same power rule as everything else.

Key idea

Powers first, calculus second, algebra third. Show each one and the method marks are safe.

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