IB Mathematics Applications & Interpretation Topic SL…

One Rule Turns The Limit Into Simple Arithmetic

Finding a gradient from the limit of chord gradients works, but doing it every time would be painfully slow.
For powers of x there is a shortcut that gives the derivative in one line, called the power rule.
Once you have it, differentiating a polynomial becomes bookkeeping rather than a limit calculation.

Deriving The Power Rule From The Difference Quotient

Start from the difference quotient for $f(x) = x^2$.
Expand the top: $$\dfrac{(x+h)^2 - x^2}{h} = \dfrac{x^2 + 2xh + h^2 - x^2}{h} = \dfrac{2xh + h^2}{h}$$
Cancel one factor of $h$: $$\dfrac{2xh + h^2}{h} = 2x + h$$
As $h \to 0$ this approaches $2x$, so the derivative of $x^2$ is $2x$.
That matches the pattern below, which holds for every integer power: $$\frac{d}{dx}\left(ax^n\right) = an\,x^{n-1}, \quad n \in \mathbb{Z}$$

Note

In The Booklet

The power rule is printed in the AI formula booklet, so you select and apply it rather than recall it.
At SL the power $n$ is an integer; fractional powers are an HL extension.

Applying The Rule Term By Term

Differentiate each term separately, because the derivative of a sum is the sum of the derivatives.
A constant multiplier stays put: the derivative of $ax^n$ is $an\,x^{n-1}$.
The derivative of a plain constant is $0$, and the derivative of $x$ is $1$.

Example

To differentiate $f(x) = 5x^3 - 2x + 7$:

Term by term: $15x^2$, then $-2$, then $0$.
So $f'(x) = 15x^2 - 2$.

Example question

Differentiate $f(x) = 5x^3 - 2x + 7$.

Solution

Bring down each power: $$5(3)x^2 = 15x^2$$
The derivative of $-2x$ is $-2$, and of $7$ is $0$.

$$f'(x) = 15x^2 - 2$$

Negative Powers: Rewrite Before You Differentiate

The rule also covers negative integer powers, but only after you rewrite fractions as powers of $x$.
For example $\dfrac{4}{x^3}$ becomes $4x^{-3}$, which you can now differentiate.

Example

To differentiate $f(x) = x^2 - \dfrac{4}{x^3}$:

Rewrite as $x^2 - 4x^{-3}$.
Differentiate: $2x + 12x^{-4}$
Tidy back to fraction form: $f'(x) = 2x + \dfrac{12}{x^4}$

Example question

Differentiate $f(x) = x^2 - \dfrac{4}{x^3}$.

Solution

Rewrite: $$x^2 - 4x^{-3}$$
Apply the power rule: $$2x - 4(-3)x^{-4} = 2x + 12x^{-4}$$

$$f'(x) = 2x + \dfrac{12}{x^4}$$

Evaluating A Derivative At A Point

To get the gradient at one value of $x$, differentiate first, then substitute the value.
Your GDC can compute a numerical derivative at a point, which is a fast independent check.

Example question

Find the gradient of $f(x) = 2x^3 - 5x$ at $x = 1$.

Solution

Differentiate: $$f'(x) = 6x^2 - 5$$
Substitute: $$f'(1) = 6(1) - 5 = 1$$
The gradient at $x = 1$ is $1$.

Exam technique

The power rule is in the booklet; the marks are for applying it correctly.
Always rewrite negative powers as $x^{-n}$ before differentiating.
At SL only integer powers appear, so you never meet $x^{1/2}$ here.
Use the GDC numerical derivative to check a value, but show the rule as your method.

Active recall

State the power rule for $\dfrac{d}{dx}(ax^n)$.
Differentiate $5x^3 - 2x + 7$.
Why must you rewrite $\dfrac{4}{x^3}$ before differentiating, and what does it become?
What is the derivative of a constant term?

One Rule Turns The Limit Into Simple Arithmetic

Finding a gradient from the limit of chord gradients works, but doing it every time would be painfully slow.
For powers of x there is a shortcut that gives the derivative in one line, called the power rule.
Once you have it, differentiating a polynomial becomes bookkeeping rather than a limit calculation.

Deriving The Power Rule From The Difference Quotient

Start from the difference quotient for $f(x) = x^2$.
Expand the top: $$\dfrac{(x+h)^2 - x^2}{h} = \dfrac{x^2 + 2xh + h^2 - x^2}{h} = \dfrac{2xh + h^2}{h}$$
Cancel one factor of $h$: $$\dfrac{2xh + h^2}{h} = 2x + h$$
As $h \to 0$ this approaches $2x$, so the derivative of $x^2$ is $2x$.
That matches the pattern below, which holds for every integer power: $$\frac{d}{dx}\left(ax^n\right) = an\,x^{n-1}, \quad n \in \mathbb{Z}$$

Note

In The Booklet

The power rule is printed in the AI formula booklet, so you select and apply it rather than recall it.
At SL the power $n$ is an integer; fractional powers are an HL extension.

Applying The Rule Term By Term

Differentiate each term separately, because the derivative of a sum is the sum of the derivatives.
A constant multiplier stays put: the derivative of $ax^n$ is $an\,x^{n-1}$.
The derivative of a plain constant is $0$, and the derivative of $x$ is $1$.

Example

To differentiate $f(x) = 5x^3 - 2x + 7$:

Term by term: $15x^2$, then $-2$, then $0$.
So $f'(x) = 15x^2 - 2$.

Example question

Differentiate $f(x) = 5x^3 - 2x + 7$.

Solution

Bring down each power: $$5(3)x^2 = 15x^2$$
The derivative of $-2x$ is $-2$, and of $7$ is $0$.

$$f'(x) = 15x^2 - 2$$

Negative Powers: Rewrite Before You Differentiate

The rule also covers negative integer powers, but only after you rewrite fractions as powers of $x$.
For example $\dfrac{4}{x^3}$ becomes $4x^{-3}$, which you can now differentiate.

Example

To differentiate $f(x) = x^2 - \dfrac{4}{x^3}$:

Rewrite as $x^2 - 4x^{-3}$.
Differentiate: $2x + 12x^{-4}$
Tidy back to fraction form: $f'(x) = 2x + \dfrac{12}{x^4}$

Example question

Differentiate $f(x) = x^2 - \dfrac{4}{x^3}$.

Solution

Rewrite: $$x^2 - 4x^{-3}$$
Apply the power rule: $$2x - 4(-3)x^{-4} = 2x + 12x^{-4}$$

$$f'(x) = 2x + \dfrac{12}{x^4}$$

Evaluating A Derivative At A Point

To get the gradient at one value of $x$, differentiate first, then substitute the value.
Your GDC can compute a numerical derivative at a point, which is a fast independent check.

Example question

Find the gradient of $f(x) = 2x^3 - 5x$ at $x = 1$.

Solution

Differentiate: $$f'(x) = 6x^2 - 5$$
Substitute: $$f'(1) = 6(1) - 5 = 1$$
The gradient at $x = 1$ is $1$.

Exam technique

The power rule is in the booklet; the marks are for applying it correctly.
Always rewrite negative powers as $x^{-n}$ before differentiating.
At SL only integer powers appear, so you never meet $x^{1/2}$ here.
Use the GDC numerical derivative to check a value, but show the rule as your method.

Active recall

State the power rule for $\dfrac{d}{dx}(ax^n)$.
Differentiate $5x^3 - 2x + 7$.
Why must you rewrite $\dfrac{4}{x^3}$ before differentiating, and what does it become?
What is the derivative of a constant term?

IB Mathematics Applications & Interpretation SL 5.3—introduction to Derivatives Notes

IB Mathematics Applications & Interpretation SL 5.3—introduction to Derivatives Notes

One Rule Turns The Limit Into Simple Arithmetic

Deriving The Power Rule From The Difference Quotient

Applying The Rule Term By Term

Negative Powers: Rewrite Before You Differentiate

Evaluating A Derivative At A Point

One Rule Turns The Limit Into Simple Arithmetic

Deriving The Power Rule From The Difference Quotient

Applying The Rule Term By Term

Negative Powers: Rewrite Before You Differentiate

Evaluating A Derivative At A Point

IB Mathematics Applications & Interpretation SL 5.3—introduction to Derivatives Notes

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

IB Mathematics Applications & Interpretation SL 5.3—introduction to Derivatives Notes

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

One Rule Turns The Limit Into Simple Arithmetic

Deriving The Power Rule From The Difference Quotient

Applying The Rule Term By Term

Negative Powers: Rewrite Before You Differentiate

Evaluating A Derivative At A Point

One Rule Turns The Limit Into Simple Arithmetic

Deriving The Power Rule From The Difference Quotient

Applying The Rule Term By Term

Negative Powers: Rewrite Before You Differentiate

Evaluating A Derivative At A Point