Basic Differentiation (AI SL)
Differentiation turns a function into its gradient function, and for IB AI SL that almost always means applying the power rule to a sum of \(x^n\) terms. This page covers the rule itself, how to handle terms written as fractions or roots, and the mistakes that come up most often. It's part of the wider Differentiation topic.
46 questions on this sub-topic.
The power rule
Covered under IB syllabus reference SL5.3. This is not in the formula booklet - it's assumed prior knowledge you're expected to apply directly.
Power rule
\(f(x) = ax^n \Rightarrow f'(x) = anx^{n-1}\)
Multiply by the exponent, then reduce the exponent by one. Applies term-by-term to any sum such as \(f(x)=ax^n+bx^{n-1}+\ldots\)
Gradient at a point
\(f'(a)\) = gradient of the curve at \(x=a\)
Differentiate first, then substitute the given \(x\)-value into the derivative - not into the original function.
Most GDCs can also evaluate a numerical derivative at a point directly - see the parent topic's GDC guidance for the exact keystrokes.
Worked examples
A curve has equation \(f(x) = x^2 - 4x + 1\).
Find the gradient of the curve at the point where \(x = 3\).
Worked solution
\(f'(x) = 2x - 4.\) M1 A1
\(f'(3) = 2(3) - 4 = 2.\) A1
A function is \(f(x) = 3x^2 + \dfrac{6}{x}.\)
(a) Write \(f(x)\) using a negative index.
(b) Find \(f'(x).\)
Worked solution
(a) \(f(x) = 3x^2 + 6x^{-1}.\) M1 A1
(b) \(f'(x) = 6x - 6x^{-2}\) M1
\(= 6x - \dfrac{6}{x^2}.\) A1
For \(y=x^2\), find \(x\) where the gradient equals \(10\).
Worked solution
\(2x=10\Rightarrow x=5.\) M1
\(x=5.\) A1
Common mistakes
- Substituting into \(f(x)\) instead of \(f'(x)\) when a gradient is asked for. "Find the gradient at \(x=3\)" means differentiate first, then substitute - substituting into the original function gives the \(y\)-value, not the gradient.
- Forgetting to rewrite roots and fractions as powers before differentiating. A term like \(\dfrac{6}{x}\) or \(\sqrt{x}\) must first become \(6x^{-1}\) or \(x^{1/2}\) - the power rule can't be applied directly to a term written any other way.
- Dropping the constant term instead of sending it to zero. Differentiating a function like \(f(x)=x^2-4x+1\) removes the constant \(+1\) entirely rather than leaving it behind, since its gradient contribution is zero.
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46 basic differentiation questions, marked instantly like the real exam.
Quick answers
What is the power rule for differentiation?
If \(f(x) = ax^n\), then \(f'(x) = anx^{n-1}\). Multiply by the exponent, then reduce the exponent by one. It applies term by term to any sum of power terms.
How do you differentiate a term with x in the denominator?
Rewrite it using a negative index first - for example \(\dfrac{6}{x}\) becomes \(6x^{-1}\) - then apply the power rule as normal.