IB Mathematics: Analysis and Approaches SL/HL — Topic 5 Calculus

Power Rule Differentiation

Turn a function into its gradient function efficiently, then use the derivative to answer gradient and rate-of-change questions.

AA SL/HL · SL 5.3

Learning goal

Differentiate integer powers and sums of powers, then use the derivative to find gradients and rates of change.

Syllabus link

IB Mathematics AA SL/HL: SL 5.3. Derivatives of \(ax^n\), \(n\in\mathbb Z\), and sums of these functions.

Big idea

Differentiation creates a new function: \(f'(x)\). Evaluating \(f'(a)\) gives the gradient of the original curve at \(x=a\).

Key relationship

\(\boxed{ax^n\longrightarrow anx^{n-1}}\). Multiply by the old power, then reduce the power by 1.

The derivative function: what are we finding?

In 5.1, \(f'(x)\) was introduced as the function that gives the gradient of \(f\) at each x-value. This lesson gives us an efficient algebraic rule for finding that derivative.

If \(y=f(x)\), then \(f'(x)\) or \(dy/dx\) tells us the tangent gradient at each input value. Once we have the derivative function, a gradient question becomes an evaluation question.

From function to tangent gradient

\[f(x)\xrightarrow{\text{differentiate}}f'(x)\xrightarrow{x=a}f'(a).\]

Curve with tangent illustrating that the derivative value is the tangent gradient — The derivative function stores the tangent gradient of the original curve.

The power rule

Power rule

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

Constant multiple

Keep the constant coefficient and differentiate the power.

\[\frac{d}{dx}(7x^4)=28x^3.\]

Sums and differences

Differentiate term by term.

\[\frac{d}{dx}(x^4-3x^2+5x)=4x^3-6x+5.\]

Simple cases, constants and negative powers

Function	Derivative	Reason
\(7x^4\)	\(28x^3\)	\(7\times4=28\), then \(4-1=3\)
\(-3x\)	\(-3\)	Think of \(x\) as \(x^1\)
\(5\)	0	A constant does not change
\(\frac8x=8x^{-1}\)	\(-8x^{-2}=-\frac8{x^2}\)	Rewrite as a power first
\(\frac6{x^3}=6x^{-3}\)	\(-18x^{-4}=-\frac{18}{x^4}\)	Negative integer power

Rewrite reciprocals before differentiating

For this syllabus point, it is often much clearer to write \(1/x^k=x^{-k}\) first. Then the same power rule applies without introducing a new formula.

Worked example 1: differentiate term by term

Differentiate \(f(x)=4x^5-3x^3+7x-11\).

\[ f'(x)=20x^4-9x^2+7. \]

The constant \(-11\) differentiates to 0.

Worked example 2: rewrite a reciprocal

Differentiate

\[y=x^2+\frac{64}{x}=x^2+64x^{-1}.\]

Now apply the power rule:

\[\boxed{\frac{dy}{dx}=2x-64x^{-2}=2x-\frac{64}{x^2}}.\]

Using the derivative to find a gradient

Worked example 3: gradients at selected points

For

\[f(x)=x^3-3x^2-4x+5,\]

first differentiate:

\[f'(x)=3x^2-6x-4.\]

Then evaluate the derivative at the required inputs:

\[f'(2)=12-12-4=-4,\qquad f'(-1)=3+6-4=5.\]

Gradients At \(x=2\), the curve has gradient \(-4\); at \(x=-1\), it has gradient 5.

Worked example 4: tangent gradient for a reciprocal expression

For \(y=x^2+\frac{64}{x}\), find the tangent gradient at \(x=4\).

From worked example 2,

\[\frac{dy}{dx}=2x-\frac{64}{x^2}.\]

Substitute \(x=4\):

\[\left.\frac{dy}{dx}\right|_{x=4}=8-\frac{64}{16}=4.\]

Answer The tangent gradient is 4.

Products and quotients of powers: simplify first

At this stage of the course, if an expression is a product or quotient of polynomial powers that can be expanded or simplified, do that algebra first. Then differentiate the resulting sum of powers.

Index laws remain useful

\[x^a x^b=x^{a+b},\qquad \frac{x^a}{x^b}=x^{a-b}.\]

Worked example 5: simplify, then differentiate

(a) \(y=x^2(4x^2+3x-5)\)

\[y=4x^4+3x^3-5x^2\quad\Rightarrow\quad y'=16x^3+9x^2-10x.\]

(b) \(y=(x^3+2x)(3x^2-1)\)

\[y=3x^5+5x^3-2x\quad\Rightarrow\quad y'=15x^4+15x^2-2.\]

(c) \(y=\frac{8x^5-3x^2+10}{x^2}\)

\[y=8x^3-3+10x^{-2}\quad\Rightarrow\quad y'=24x^2-20x^{-3}=24x^2-\frac{20}{x^3}.\]

IB-style worked examples

Worked example A: equation of a tangent

Let \(f(x)=x^3-2x^2+5x-7\). Find the equation of the tangent at \(x=1\).

Step 1: differentiate.

\[f'(x)=3x^2-4x+5.\]

Step 2: find the point and gradient.

\[f(1)=1-2+5-7=-3,\qquad f'(1)=3-4+5=4.\]

Step 3: use point-gradient form.

\[y+3=4(x-1)\quad\Rightarrow\quad \boxed{y=4x-7}.\]

Worked example B: determine an unknown coefficient

Suppose \(f(x)=3x^{-2}+bx^{-1}+c\), and the gradient at \(x=4\) is \(5/32\). Find \(b\), then determine whether the function is increasing at \(x=5\).

\[f'(x)=-6x^{-3}-bx^{-2}.\]

Use \(f'(4)=5/32\):

\[-\frac6{64}-\frac b{16}=\frac5{32}.\]

Multiplying by 32 gives \(-3-2b=5\), so \(b=-4\).

Then

\[f'(5)=-\frac6{125}+\frac4{25}=\frac{14}{125}>0.\]

Therefore the function is increasing at \(x=5\).

Worked example C: rate of change in context

A liquid volume is modelled by

\[V(t)=800+24t-1.5t^2+0.02t^3,\qquad 0\le t\le20.\]

Differentiate:

\[V'(t)=24-3t+0.06t^2.\]

At \(t=5\),

\[V'(5)=24-15+1.5=10.5\text{ L/min}.\]

For a stationary volume, solve \(V'(t)=0\):

\[24-3t+0.06t^2=0\quad\Rightarrow\quad t=10\text{ or }40.\]

Only \(t=10\) is in the model domain. At that instant the volume is momentarily neither increasing nor decreasing.

Common mistakes

Check these before finalising a derivative

Do not keep the same exponent. The exponent must decrease by 1.
Do not forget to multiply by the old exponent. \(x^5\) differentiates to \(5x^4\), not \(x^4\).
Do not differentiate a reciprocal as written if a negative power is clearer. Rewrite \(1/x^3=x^{-3}\).
Constants differentiate to zero.
Simplify expandable products and quotients first. The power rule can then be applied term by term.
A derivative value needs interpretation. In a context, include the units and explain whether the quantity is increasing or decreasing.

Practice

A. Differentiate

\(y=x^7\)
\(y=5x^3-4x+9\)
\(y=12x^5-7x^2+3\)
\(y=x^{-3}\)
\(y=\frac6{x^2}\)
\(y=4x^2-\frac{10}{x}+7\)

B. Simplify first, then differentiate

\(y=x^2(3x^3-2x+1)\)
\(y=(x^2-3x)(4x^2+1)\)
\(y=\frac{4x^6-3x^3+8}{x^2}\)
\(y=\frac{5x^4+2x^2-1}{x}\)

C. Gradients, tangents and context

For \(f(x)=x^3-5x^2+6x\), find \(f'(2)\).
For \(y=x^2+\frac{25}{x}\), find the tangent gradient at \(x=5\).
Find the equation of the tangent to \(f(x)=2x^3-x\) at \(x=1\).
For \(f(x)=kx^3-4x\), the gradient at \(x=2\) is 20. Find \(k\).
A height is modelled by \(h(t)=2t^3-9t^2+12t+5\). Find \(h'(t)\) and \(h'(3)\).
For \(f(x)=x^3-3x^2-9x+1\), find the x-values where the tangent is horizontal.

Concise answer key

1. \(7x^6\)

2. \(15x^2-4\)

3. \(60x^4-14x\)

4. \(-3x^{-4}\)

5. \(-12/x^3\)

6. \(8x+10/x^2\)

7. \(15x^4-6x^2+2x\)

8. \(16x^3-36x^2+2x-3\)

9. \(16x^3-3-16/x^3\)

10. \(15x^2+2+1/x^2\)

11. \(-2\)

12. 9

13. \(y=5x-4\)

14. \(k=2\)

15. \(h'=6t^2-18t+12\), \(h'(3)=12\)

16. \(x=-1,3\)

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