This note gives the power rule for differentiating terms of the form \(ax^n\), where \(n\) is a positive or negative integer, and shows how to differentiate sums of such terms. It also shows how to rewrite fractions and brackets so that the rule can be used. These skills are used in almost every later calculus question.
Key ideas
The power rule
To differentiate \(ax^n\): multiply by the power, then reduce the power by 1.
\[ \frac{d}{dx}\left(ax^n\right)=anx^{n-1}. \]This works for every integer \(n\), including negative ones. For example, \(\dfrac{d}{dx}(x^5)=5x^4\) and \(\dfrac{d}{dx}(x^{-2})=-2x^{-3}\).
Special cases
- A constant has zero gradient: \(\dfrac{d}{dx}(c)=0\).
- \(\dfrac{d}{dx}(ax)=a\), since \(ax=ax^1\) and \(x^0=1\).
Sums, differences and multiples
You differentiate a polynomial one term at a time. A constant multiplier stays in place, and the terms are added or subtracted as before:
\[ \frac{d}{dx}\bigl(f(x)\pm g(x)\bigr)=f'(x)\pm g'(x),\qquad \frac{d}{dx}\bigl(kf(x)\bigr)=kf'(x). \]Rewriting before differentiating
The power rule only applies to terms written as a number times a power of \(x\). Often you must rearrange first.
- Reciprocals: \(\dfrac{1}{x^n}=x^{-n}\). So \(\dfrac{5}{x^3}=5x^{-3}\).
- Division by a single term: split the fraction. \(\dfrac{2x^3-5x+4}{x^2}=2x-5x^{-1}+4x^{-2}\).
- Brackets: expand them. \((2x-1)(x+3)^2\) must be multiplied out to a polynomial before using the rule.
Take care with a number in the denominator. The fraction \(\dfrac{3}{2x}\) equals \(\dfrac32x^{-1}\), because the 2 stays with the number, not with \(x\). Its derivative is \(-\dfrac32x^{-2}=-\dfrac{3}{2x^2}\).
Why the rule works
In the previous note, expanding \((x+h)^n\) gave \(x^n+nx^{n-1}h+\dots\), where every later term contains \(h^2\) or a higher power of \(h\). After dividing by \(h\) and letting \(h\to 0\), only \(nx^{n-1}\) is left. The rule then extends to negative integer powers, and you may use it for all integers. You do not need to prove it again each time. Just apply it carefully.
A routine for every question
- Expand brackets and split fractions so that every term is a number times a power of \(x\).
- Differentiate term by term: multiply by the power, then subtract 1 from the power.
- If the question asks for a gradient, substitute the \(x\)-value into \(f'(x)\) only at the end.
- If the question gives a gradient, set \(f'(x)\) equal to that number and solve.
Leave your final answer in the form that suits the question. If it asks for a gradient at a point, give a number. If it asks for \(\dfrac{dy}{dx}\), either negative powers or fractions are fine, as long as the signs are right. Always check by substituting a simple value, such as \(x=1\), into both the unsimplified and simplified derivative, because it quickly exposes slips in signs.
Using the derivative
Once you have \(f'(x)\), you can find the gradient at a point by substitution, find where the gradient has a given value by solving an equation, and find where \(f'(x)=0\) for stationary points. The derivative may also be given a context, for example the rate at which a population or a cost is changing.
A word about context
If \(C(n)\) is the cost in euros of producing \(n\) items, then \(C'(n)\) is the rate at which cost changes with the number of items, in euros per item. If \(V(t)\) is the volume of water in a tank in litres after \(t\) minutes, then \(V'(t)\) is in litres per minute. A negative value shows that the quantity is falling.
Notation reminder
If \(y=x^3\), write \(\dfrac{dy}{dx}=3x^2\). If \(f(x)=x^3\), write \(f'(x)=3x^2\). Use whichever notation the question uses.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\dfrac{d}{dx}(ax^n)=anx^{n-1}\) | Power rule, for any integer \(n\). |
| \(\dfrac{d}{dx}(c)=0\) | The derivative of a constant term is zero. |
| \(\dfrac{d}{dx}(ax)=a\) | A term in \(x\) only differentiates to its coefficient. |
| \(\dfrac{1}{x^n}=x^{-n}\) | Rewrite fractions with \(x\) in the denominator before differentiating. |
| \(f'(x)\pm g'(x)\) | Differentiate sums and differences term by term. |
Worked examples
Example 1 Given \(f(x)=4x^3-7x^2+5x-9\), find \(f'(x)\) and calculate \(f'(2)\).
Differentiate each term: \(4x^3\to 12x^2\), \(-7x^2\to -14x\), \(5x\to 5\), \(-9\to 0\).
So \(f'(x)=12x^2-14x+5\).
Substitute \(x=2\): \(f'(2)=12(4)-14(2)+5=48-28+5=25\).
Answer: \(f'(x)=12x^2-14x+5\); \(f'(2)=25\).
Example 2 Differentiate \(y=\dfrac{2x^3-5x+4}{x^2}\) and find the gradient when \(x=2\).
Split the fraction and write each term as a power: \(y=2x-5x^{-1}+4x^{-2}\).
Differentiate: \(\dfrac{dy}{dx}=2+5x^{-2}-8x^{-3}\).
The term \(-5x^{-1}\) gives \((-5)(-1)x^{-2}=5x^{-2}\), and \(4x^{-2}\) gives \((4)(-2)x^{-3}=-8x^{-3}\).
Rewrite as \(\dfrac{dy}{dx}=2+\dfrac{5}{x^2}-\dfrac{8}{x^3}\).
At \(x=2\): \(2+\dfrac54-\dfrac88=2+1.25-1=2.25\).
Answer: \(\dfrac{dy}{dx}=2+\dfrac{5}{x^2}-\dfrac{8}{x^3}\); the gradient at \(x=2\) is \(2.25\).
Example 3 The curve \(y=(2x-1)(x+3)^2\) has gradient 12 at two points. Find the \(x\)-coordinates of these points.
Expand first: \((x+3)^2=x^2+6x+9\).
\((2x-1)(x^2+6x+9)=2x^3+12x^2+18x-x^2-6x-9=2x^3+11x^2+12x-9\).
Differentiate: \(\dfrac{dy}{dx}=6x^2+22x+12\).
Set the gradient equal to 12: \(6x^2+22x+12=12\), so \(6x^2+22x=0\).
Factorise: \(2x(3x+11)=0\), so \(x=0\) or \(x=-\dfrac{11}{3}\).
Answer: \(x=0\) and \(x=-\dfrac{11}{3}\) (about \(-3.67\)).
Common errors
- Common error: forgetting to subtract 1 from the power, or subtracting 1 from the coefficient instead. For \(5x^4\), the derivative is \(20x^3\), not \(5x^3\) or \(4x^3\).
- Common error: getting negative powers wrong. The derivative of \(x^{-3}\) is \(-3x^{-4}\), not \(-3x^{-2}\). Subtracting 1 from \(-3\) gives \(-4\).
- Common error: differentiating a fraction such as \(\dfrac{3}{x^2}\) by differentiating the top and bottom separately. Rewrite it as \(3x^{-2}\) first.
- Common error: keeping a constant term. The derivative of \(-9\) is 0, not \(-9\).
- Common error: using the power rule on a bracket such as \((2x-1)(x+3)^2\) without expanding. Multiply out first.
Exam tip: when a question says "find the gradient" after a long rearrangement, do the substitution last and show the numbers in the line before the answer. Write each rewritten term as a number times \(x\) to a power on its own line before you differentiate. Check each sign when the power is negative.
Check yourself
- Differentiate \(y=5x^4-2x^3+7x-1\).
Answer
\(\dfrac{dy}{dx}=20x^3-6x^2+7\). - Differentiate \(y=\dfrac{4}{x^2}-\dfrac{3}{x}\).
Answer
Rewrite as \(y=4x^{-2}-3x^{-1}\). Then \(\dfrac{dy}{dx}=-8x^{-3}+3x^{-2}=-\dfrac{8}{x^3}+\dfrac{3}{x^2}\). - Given \(f(x)=x^2(x-5)\), find \(f'(2)\).
Answer
Expand: \(f(x)=x^3-5x^2\). So \(f'(x)=3x^2-10x\) and \(f'(2)=12-20=-8\). - Find the \(x\)-values at which the curve \(y=x^3-12x\) has zero gradient.
Answer
\(\dfrac{dy}{dx}=3x^2-12=0\), so \(x^2=4\) and \(x=2\) or \(x=-2\).