Year 12 Maths - Methods (Unit 3 & Unit 4) Differentiation

Differentiating x^n where n is a negative integer

Q: How do you differentiate 1 over x?

Rewrite 1 over x as x to the power minus 1, then use the power rule. Differentiating gives minus 1 times x to the power minus 2, which is minus 1 over x squared.

Q: Does the power rule work for negative powers?

Yes. The power rule d/dx of x to the n equals n times x to the power n minus 1 holds for every integer n, including negative integers. For example d/dx of x to the minus 3 equals minus 3 times x to the minus 4.

Q: How do you rewrite a fraction like 3 over x cubed before differentiating?

Move the power of x into the numerator with a negative index: 3 over x cubed equals 3 times x to the minus 3. Then differentiate to get 3 times minus 3 times x to the minus 4, which is minus 9 over x to the fourth.

Q: How do you differentiate a sum of reciprocal terms?

Rewrite each term as a power of x, differentiate term by term with the power rule, then convert any negative indices back to fractions. For example x squared minus 4 over x differentiates to 2x plus 4 over x squared.

Q: How do you find the gradient of y equals 1 over x at a point?

Differentiate to get dy/dx equals minus 1 over x squared, then substitute the x value of the point. For y equals 2 over x at the point (1,2) the gradient is minus 2 and the tangent line is y equals minus 2x plus 4.

Q: Why is the derivative undefined at x equals 0?

A term like 1 over x is not defined at x equals 0, so its derivative minus 1 over x squared is also undefined there. The domain of these functions and their derivatives excludes x equals 0.

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Theory

The power rule \(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\) extends to negative integer powers, so reciprocal terms such as \(\dfrac{1}{x}\), \(\dfrac{1}{x^{2}}\) and \(\dfrac{3}{x^{3}}\) are first rewritten as \(x^{-1}\), \(x^{-2}\) and \(3x^{-3}\) before differentiating. Each term is differentiated separately and written back in fraction form; the function and its derivative are undefined at \(x=0\).

The power rule \(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\) holds for every integer \(n\), not just positive ones. When \(n\) is a negative integer the same procedure applies — multiply by the index and reduce the index by one — so, for example, \(\dfrac{d}{dx}x^{-2}=-2x^{-3}\).

To use the rule on a fraction, first rewrite it as a power of \(x\) using the index law \(\dfrac{1}{x^{k}}=x^{-k}\). Thus \(\dfrac{1}{x}=x^{-1}\), \(\dfrac{1}{x^{2}}=x^{-2}\), and a constant on top is kept as a coefficient: \(\dfrac{3}{x^{3}}=3x^{-3}\). After differentiating, negative indices are usually converted back to fractions so the answer matches the original form.

A sum or difference of such terms is differentiated term by term, and a constant multiple is carried through: \(\dfrac{d}{dx}\big(a\,x^{n}\big)=a\,n\,x^{n-1}\). Because a term like \(\dfrac{1}{x}\) has no value at \(x=0\), the domain excludes \(x=0\), and the derivative gives the gradient only at points where \(x\neq 0\).

Key idea. \(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\) for any integer \(n\). Rewrite \(\dfrac{a}{x^{k}}\) as \(a\,x^{-k}\) before differentiating, then apply the power rule — and remember \(x\neq 0\).

\(y=\dfrac{1}{x}=x^{-1}\): the gradient is \(-\dfrac{1}{x^{2}}\), so at \((1,1)\) it is \(-1\) and the tangent is \(y=-x+2\)

\(y=\dfrac{1}{x^{2}}=x^{-2}\) is positive on both branches and undefined at \(x=0\); its derivative is \(-\dfrac{2}{x^{3}}\)

The power rule — valid for every integer \(n\), including negative integers:

\[\frac{d}{dx}x^{n}=n\,x^{n-1}\qquad (n\in\mathbb{Z})\]

\frac{d}{d x} x^{n} = n x^{n - 1}

Rewrite reciprocals as negative powers before differentiating, then convert back:

\[\frac{a}{x^{k}}=a\,x^{-k}\qquad \frac{d}{dx}\big(a\,x^{-k}\big)=-a\,k\,x^{-k-1}=-\frac{ak}{x^{k+1}}\]

\frac{d}{d x} x^{- k} = - k x^{- k - 1}

The two standard cases, written in fraction form:

\[\frac{d}{dx}\!\left(\frac{1}{x}\right)=-\frac{1}{x^{2}}\qquad \frac{d}{dx}\!\left(\frac{1}{x^{2}}\right)=-\frac{2}{x^{3}}\]

\frac{d}{d x} \frac{1}{x} = - \frac{1}{x^{2}}

Gradient at a point. Substitute the \(x\)-value (with \(x\neq 0\)) into the derivative. For \(y=\dfrac{2}{x}=2x^{-1}\), \(\dfrac{dy}{dx}=-\dfrac{2}{x^{2}}\), so at \(x=1\) the gradient is \(-2\).

How to differentiate a term with a negative index

Rewrite as a power of \(x\). Use \(\dfrac{a}{x^{k}}=a\,x^{-k}\) so every term looks like \(a\,x^{n}\) — for example \(\dfrac{3}{x^{3}}=3x^{-3}\).
Apply the power rule to each term. Multiply by the index and reduce the index by \(1\): \(\dfrac{d}{dx}\big(a\,x^{n}\big)=a\,n\,x^{n-1}\). Take care with signs when \(n\) is negative.
Convert back to fractions. Rewrite negative indices as reciprocals, e.g. \(-9x^{-4}=-\dfrac{9}{x^{4}}\).
Evaluate if required. For a gradient, substitute the \(x\)-value (\(x\neq 0\)); for a tangent, use \(y-y_{1}=m(x-x_{1})\) with that gradient and the point.

Sign check. For a negative index the new index \(n-1\) is more negative: \(\dfrac{d}{dx}x^{-2}=-2x^{-3}\), not \(-2x^{-1}\). Reduce the index — do not raise it.

Example 1 — differentiate \(\dfrac{1}{x}\)

Differentiate \(y=\dfrac{1}{x}\).

Solution

Rewrite as a power of \(x\):

\(y\)

\(=\)

\(x^{-1}\)

Apply the power rule \(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\):

\(\dfrac{dy}{dx}\)	\(=\)	\(-1\,x^{-2}\)
	\(=\)	\(-\dfrac{1}{x^{2}}\)

\(\therefore\) \(\dfrac{dy}{dx}=-\dfrac{1}{x^{2}}\)

- \frac{1}{x^{2}}

Example 2 — a constant over a power

Differentiate (a) \(y=\dfrac{1}{x^{2}}\) and (b) \(y=\dfrac{3}{x^{3}}\).

Solution

(a) \(y=\dfrac{1}{x^{2}}\)

Rewrite, then differentiate:

\(y\)	\(=\)	\(x^{-2}\)
\(\dfrac{dy}{dx}\)	\(=\)	\(-2x^{-3}=-\dfrac{2}{x^{3}}\)

\(\therefore\) \(\dfrac{dy}{dx}=-\dfrac{2}{x^{3}}\)

(b) \(y=\dfrac{3}{x^{3}}\)

Keep the \(3\) as a coefficient, rewrite, then differentiate:

\(y\)	\(=\)	\(3x^{-3}\)
\(\dfrac{dy}{dx}\)	\(=\)	\(3\times(-3)\,x^{-4}\)
	\(=\)	\(-9x^{-4}=-\dfrac{9}{x^{4}}\)

\(\therefore\) \(\dfrac{dy}{dx}=-\dfrac{9}{x^{4}}\)

- \frac{9}{x^{4}}

Example 3 — a sum of terms

Differentiate \(y=x^{2}-\dfrac{4}{x}+\dfrac{2}{x^{3}}\).

Solution

Rewrite every term as a power of \(x\):

\(y\)

\(=\)

\(x^{2}-4x^{-1}+2x^{-3}\)

Differentiate term by term:

\(\dfrac{dy}{dx}\)	\(=\)	\(2x-4(-1)x^{-2}+2(-3)x^{-4}\)
	\(=\)	\(2x+4x^{-2}-6x^{-4}\)

Write negative indices back as fractions:

\(\dfrac{dy}{dx}\)

\(=\)

\(2x+\dfrac{4}{x^{2}}-\dfrac{6}{x^{4}}\)

\(\therefore\) \(\dfrac{dy}{dx}=2x+\dfrac{4}{x^{2}}-\dfrac{6}{x^{4}}\)

2 x + \frac{4}{x^{2}} - \frac{6}{x^{4}}

Example 4 — gradient and tangent

For \(y=\dfrac{2}{x}\), find the gradient at \((1,2)\) and the equation of the tangent there.

Solution

Rewrite and differentiate:

\(y\)	\(=\)	\(2x^{-1}\)
\(\dfrac{dy}{dx}\)	\(=\)	\(-2x^{-2}=-\dfrac{2}{x^{2}}\)

Gradient — substitute \(x=1\):

\(m\)

\(=\)

\(-\dfrac{2}{1^{2}}=-2\)

Tangent — use \(y-y_{1}=m(x-x_{1})\) at \((1,2)\):

\(y-2\)	\(=\)	\(-2(x-1)\)
\(y\)	\(=\)	\(-2x+4\)

\(\therefore\) gradient \(=-2\); tangent: \(y=-2x+4\)

y = - 2 x + 4

Common pitfalls

Reduce the index, do not raise it. For a negative index \(n-1\) is more negative: \(\dfrac{d}{dx}x^{-2}=-2x^{-3}\), never \(-2x^{-1}\).

Rewrite before you differentiate. The power rule applies to \(x^{-2}\), not to \(\dfrac{1}{x^{2}}\) as it stands. Convert the fraction to a negative power first, then differentiate.

Mind the domain. Terms like \(\dfrac{1}{x}\) and their derivatives are undefined at \(x=0\), so a gradient can only be found where \(x\neq 0\).

Frequently asked questions

How do you differentiate 1 over x?

Rewrite \(\dfrac{1}{x}\) as \(x^{-1}\), then use the power rule: \(\dfrac{d}{dx}x^{-1}=-1\,x^{-2}=-\dfrac{1}{x^{2}}\).

Does the power rule work for negative powers?

Yes — \(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\) holds for every integer \(n\). For example \(\dfrac{d}{dx}x^{-3}=-3x^{-4}\).

How do you rewrite a fraction like 3 over x cubed before differentiating?

Move the power into the numerator with a negative index: \(\dfrac{3}{x^{3}}=3x^{-3}\). Then \(\dfrac{dy}{dx}=3(-3)x^{-4}=-\dfrac{9}{x^{4}}\).

How do you differentiate a sum of reciprocal terms?

Rewrite each term as a power of \(x\), differentiate term by term, then convert back. For instance \(x^{2}-\dfrac{4}{x}\) gives \(2x+\dfrac{4}{x^{2}}\).

How do you find the gradient of y equals 1 over x at a point?

Differentiate to get \(\dfrac{dy}{dx}=-\dfrac{1}{x^{2}}\), then substitute the \(x\)-value. For \(y=\dfrac{2}{x}\) at \((1,2)\) the gradient is \(-2\) and the tangent is \(y=-2x+4\).

Why is the derivative undefined at x equals 0?

A term like \(\dfrac{1}{x}\) has no value at \(x=0\), so its derivative \(-\dfrac{1}{x^{2}}\) is undefined there too; the domain excludes \(x=0\).

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Differentiating x^n where n is a negative integer

📖 Theory

How to differentiate a term with a negative index

Common pitfalls

Frequently asked questions

Theory