Differentiation Part 1 — Mathslive.ie

Section 1 of 13

Rate of Change

Differentiation is about rates of change. How fast is one thing changing compared to another?

Everyday rates you already know:

Warmer day ⇒ rate at which ice melts

Speed ⇒ rate at which distance changes with time

Heart ⇒ tempo ⇒ beats per minute (a ratio)

In maths we write $y = f(x)$ and ask: what is the rate of change of $y$ with a change in $x$?

Learning is a function of time and quality — but on a graph we just take $y$ as a function of $x$.

Section 2 of 13

First Principles — The Formula

Pick a curve $y = f(x)$. Pick two points on it:

Point 1: $(x,\,f(x))$ — coords $(x_1,\,y_1)$

Point 2: $(x+h,\,f(x+h))$ — coords $(x_2,\,y_2)$

The line joining them has slope:

$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{f(x+h) - f(x)}{h}$

Here $h = x_2 - x_1$ is a small change in $x$.

Now let $h$ get smaller and smaller — the line between the two points becomes the tangent at $(x,\,f(x))$. That tangent's slope is $\tan\theta$ and is what we call $\dfrac{dy}{dx}$.

First Principles — Must learn

$\dfrac{dy}{dx} = \displaystyle\lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$

$=\;$ change in $y$ over change in $x$ $\;=\;$ slope of tangent $\;=\;\tan\theta$

Section 3 of 13

From First Principles: $f(x) = 3x + 1$

Differentiate $f(x) = 3x + 1$ from first principles.

Step 1. Write out $f(x)$ and $f(x+h)$:

$f(x) = 3x + 1$

$f(x+h) = 3(x+h) + 1 = 3x + 3h + 1$

Step 2. Subtract — find $f(x+h) - f(x)$:

$f(x+h) - f(x) = (3x + 3h + 1) - (3x + 1)$

$\hphantom{f(x+h) - f(x)} = 3h$

Step 3. Divide by $h$:

$\dfrac{f(x+h) - f(x)}{h} = \dfrac{3h}{h} = 3$

Step 4. Take the limit as $h \to 0$:

$\dfrac{dy}{dx} = \displaystyle\lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} = 3$

Makes sense — $y = 3x + 1$ is a straight line of slope $3$, so its rate of change is $3$ everywhere.

Section 4 of 13

From First Principles: $f(x) = x^{2}$

Differentiate $f(x) = x^{2}$ from first principles.

$f(x) = x^{2}$

$f(x+h) = (x+h)^{2} = x^{2} + 2xh + h^{2}$

Subtract:

$f(x+h) - f(x) = 2xh + h^{2}$

Divide by $h$:

$\dfrac{f(x+h) - f(x)}{h} = \dfrac{2xh + h^{2}}{h} = 2x + h$

Take the limit — let $h \to 0$:

$\displaystyle\lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} = 2x$

So if $y = x^{2}$, then $\dfrac{dy}{dx} = 2x$. Hold onto that — it's the first hint of the rule.

Section 5 of 13

From First Principles: $f(x) = 3x^{2} - 7x + 1$

Same routine — write, expand, subtract, divide, limit.

$f(x) = 3x^{2} - 7x + 1$

$f(x+h) = 3(x+h)^{2} - 7(x+h) + 1$

$\hphantom{f(x+h)} = 3(x^{2} + 2xh + h^{2}) - 7x - 7h + 1$

$\hphantom{f(x+h)} = 3x^{2} + 6xh + 3h^{2} - 7x - 7h + 1$

Subtract $f(x)$:

$f(x+h) - f(x) = 6xh + 3h^{2} - 7h$

Divide by $h$:

$\dfrac{f(x+h) - f(x)}{h} = 6x + 3h - 7$

Limit as $h \to 0$:

$\dfrac{dy}{dx} = 6x - 7$

Section 6 of 13

From First Principles: $f(x) = (2x - 3)^{2}$

Multiply out first.

$f(x) = (2x - 3)^{2} = 4x^{2} - 12x + 9$

$f(x+h) = 4(x+h)^{2} - 12(x+h) + 9$

$\hphantom{f(x+h)} = 4(x^{2} + 2xh + h^{2}) - 12x - 12h + 9$

$\hphantom{f(x+h)} = 4x^{2} + 8xh + 4h^{2} - 12x - 12h + 9$

$f(x+h) - f(x) = 8xh + 4h^{2} - 12h$

$\dfrac{f(x+h) - f(x)}{h} = 8x + 4h - 12$

$\displaystyle\lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} = 8x - 12$

You try

Differentiate $f(x) = 5x^{2} + 2$ from first principles.

Find $f(x+h)$, subtract $f(x)$, divide by $h$, then let $h \to 0$.

$f(x+h) = 5(x+h)^{2} + 2 = 5x^{2} + 10xh + 5h^{2} + 2$

$f(x+h) - f(x) = 10xh + 5h^{2}$

$\dfrac{f(x+h) - f(x)}{h} = 10x + 5h$

$\dfrac{dy}{dx} = 10x$

$10x$

You try

Differentiate $f(x) = (3x + 2)^{2}$ from first principles.

Multiply out first — get $9x^{2} + 12x + 4$.

$f(x) = 9x^{2} + 12x + 4$

$f(x+h) = 9(x+h)^{2} + 12(x+h) + 4 = 9x^{2} + 18xh + 9h^{2} + 12x + 12h + 4$

$f(x+h) - f(x) = 18xh + 9h^{2} + 12h$

$\dfrac{f(x+h) - f(x)}{h} = 18x + 9h + 12$

$\dfrac{dy}{dx} = 18x + 12$

$18x + 12$

Section 7 of 13

The Rule

Look at the pattern from first principles:

$y = x^{2} \;\;\;\Rightarrow\;\;\; \dfrac{dy}{dx} = 2x$

$y = 3x^{2} \;\;\;\Rightarrow\;\;\; \dfrac{dy}{dx} = 6x$

$y = 4x^{2} \;\;\;\Rightarrow\;\;\; \dfrac{dy}{dx} = 8x$

In every case: the power dropped by 1, and the old power became a multiplier out front.

The Rule — Must learn

$y = x^{n} \;\;\;\Rightarrow\;\;\; \dfrac{dy}{dx} = n\,x^{\,n-1}$

Multiply by the power and reduce the power by 1.

Quick examples:

(i) $y = x^{3}$

$\dfrac{dy}{dx} = 3x^{2}$

(ii) $y = 5x^{6}$

$\dfrac{dy}{dx} = 30x^{5}$

(iii) $y = x^{3} + 5x^{2}$

$\dfrac{dy}{dx} = 3x^{2} + 10x$

(iv) $y = 2x^{3} + 5x^{2} + 7x + 9$

$\dfrac{dy}{dx} = 6x^{2} + 10x + 7$

Each term differentiated separately. The constant $9$ disappears (its rate of change is $0$).

(v) $y = 10x$

$y = 10x^{1}$

$\dfrac{dy}{dx} = 10x^{0} = 10$

(vi) $y = 3x^{0}$

$y = 3$ (since $x^{0} = 1$)

$\dfrac{dy}{dx} = 0$

Section 8 of 13

The Rule — Your Turn

You try

Differentiate: $y = x^{7}$

Multiply by the power, reduce the power by 1.

$\dfrac{dy}{dx} = 7x^{6}$

$7x^{6}$

You try

Differentiate: $y = 4x^{5} + 3x^{2} - 8x + 11$

Differentiate term by term. Constant $11$ goes to $0$.

$4x^{5} \to 20x^{4}$

$3x^{2} \to 6x$

$-8x \to -8$

$11 \to 0$

$\dfrac{dy}{dx} = 20x^{4} + 6x - 8$

$20x^{4} + 6x - 8$

You try

Differentiate: $y = 6x^{10} - 2x^{4} + 5$

Term by term — three terms here.

$6x^{10} \to 60x^{9}$

$-2x^{4} \to -8x^{3}$

$5 \to 0$

$\dfrac{dy}{dx} = 60x^{9} - 8x^{3}$

$60x^{9} - 8x^{3}$

Section 9 of 13

Standard Form: Multiply Out First

The Rule only works on standard form:

Standard form

$y = x^{a} + x^{b} + x^{c} + \,\ldots$

A sum of powers of $x$ — like the rows in the maths tables.

If you see brackets, multiply them out before differentiating.

(i) $y = (2x + 3)(x + 5)$

$y = 2x^{2} + 10x + 3x + 15$

$y = 2x^{2} + 13x + 15$

$\dfrac{dy}{dx} = 4x + 13$

You try

Differentiate: $y = (x + 4)(2x - 1)$

Multiply out first, then differentiate.

$y = 2x^{2} - x + 8x - 4$

$y = 2x^{2} + 7x - 4$

$\dfrac{dy}{dx} = 4x + 7$

$4x + 7$

You try

Differentiate: $y = (3x - 2)(x + 6)$

Multiply out — middle terms are $+18x$ and $-2x$.

$y = 3x^{2} + 18x - 2x - 12$

$y = 3x^{2} + 16x - 12$

$\dfrac{dy}{dx} = 6x + 16$

$6x + 16$

Section 10 of 13

Standard Form: Negative Powers

If $x$ is on the bottom, bring it up using the index rule:

Index rule — Must learn

$\dfrac{1}{a^{p}} = a^{-p}$

(i) $y = \dfrac{1}{x^{3}}$

$y = x^{-3}$

$\dfrac{dy}{dx} = -3x^{-4}$

$\dfrac{dy}{dx} = -\dfrac{3}{x^{4}}$

When the power was $-3$, reducing by 1 gives $-4$ (not $-2$). Be careful.

(ii) $y = \dfrac{1}{x} + \dfrac{5}{x^{2}}$

$y = x^{-1} + 5x^{-2}$

$\dfrac{dy}{dx} = -x^{-2} - 10x^{-3}$

$\dfrac{dy}{dx} = -\dfrac{1}{x^{2}} - \dfrac{10}{x^{3}}$

You try

Differentiate: $y = \dfrac{4}{x^{2}}$

Write it as $4x^{-2}$ first.

$y = 4x^{-2}$

$\dfrac{dy}{dx} = -8x^{-3}$

$\dfrac{dy}{dx} = -\dfrac{8}{x^{3}}$

$-\dfrac{8}{x^{3}}$

You try

Differentiate: $y = \dfrac{3}{x} - \dfrac{2}{x^{4}}$

Rewrite as $3x^{-1} - 2x^{-4}$.

$y = 3x^{-1} - 2x^{-4}$

$\dfrac{dy}{dx} = -3x^{-2} + 8x^{-5}$

$\dfrac{dy}{dx} = -\dfrac{3}{x^{2}} + \dfrac{8}{x^{5}}$

$-\dfrac{3}{x^{2}} + \dfrac{8}{x^{5}}$

Section 11 of 13

Standard Form: Roots

Index rule — Must learn

$\sqrt{x} = x^{\,\frac{1}{2}}$

(i) $y = \sqrt{x}$

$y = x^{\,\frac{1}{2}}$

$\dfrac{dy}{dx} = \dfrac{1}{2}\,x^{-\frac{1}{2}}$

$\dfrac{dy}{dx} = \dfrac{1}{2} \cdot \dfrac{1}{x^{\,\frac{1}{2}}}$

$\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$

(ii) $y = \dfrac{1}{\sqrt{x}}$

$y = \dfrac{1}{x^{\,\frac{1}{2}}} = x^{-\frac{1}{2}}$

$\dfrac{dy}{dx} = -\dfrac{1}{2}\,x^{-\frac{3}{2}}$

$\dfrac{dy}{dx} = -\dfrac{1}{2} \cdot \dfrac{1}{x^{\,\frac{3}{2}}} = -\dfrac{1}{2(\sqrt{x})^{3}}$

$\dfrac{dy}{dx} = -\dfrac{1}{2x\sqrt{x}}$

Used: $x^{\,\frac{3}{2}} = (\sqrt{x})^{3} = \sqrt{x}\,\sqrt{x}\,\sqrt{x} = x\sqrt{x}$.

You try

Differentiate: $y = \sqrt{x} + \dfrac{1}{x^{2}}$

Rewrite as $x^{\,\frac{1}{2}} + x^{-2}$.

$y = x^{\,\frac{1}{2}} + x^{-2}$

$\dfrac{dy}{dx} = \dfrac{1}{2}x^{-\frac{1}{2}} - 2x^{-3}$

$\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}} - \dfrac{2}{x^{3}}$

$\dfrac{1}{2\sqrt{x}} - \dfrac{2}{x^{3}}$

You try

Differentiate: $y = 4\sqrt{x}$

Rewrite as $4x^{\,\frac{1}{2}}$.

$y = 4x^{\,\frac{1}{2}}$

$\dfrac{dy}{dx} = 4 \cdot \dfrac{1}{2}\,x^{-\frac{1}{2}} = 2x^{-\frac{1}{2}}$

$\dfrac{dy}{dx} = \dfrac{2}{\sqrt{x}}$

$\dfrac{2}{\sqrt{x}}$

Section 12 of 13

Standard Form: Factorise or Split

When you see a fraction, two tricks turn it into standard form:

1. If the top factors and shares a bracket with the bottom — factorise and cancel.

2. If the bottom is a single power of $x$ — split the fraction across the top.

(i) Factorise: $y = \dfrac{9x^{2} - 25}{3x - 5}$

Top is a difference of two squares: $9x^{2} - 25 = (3x - 5)(3x + 5)$.

$y = \dfrac{(3x - 5)(3x + 5)}{3x - 5}$

$y = 3x + 5$

$\dfrac{dy}{dx} = 3$

(ii) Split: $y = \dfrac{x^{2} + 3x + 5}{x}$

$y = \dfrac{x^{2}}{x} + \dfrac{3x}{x} + \dfrac{5}{x}$

$y = x + 3 + \dfrac{5}{x}$

$y = x + 3 + 5x^{-1}$

$\dfrac{dy}{dx} = 1 - 5x^{-2}$

$\dfrac{dy}{dx} = 1 - \dfrac{5}{x^{2}}$

(iii) Split: $y = \dfrac{x^{2} + 7x + 5}{x}$

$y = x + 7 + \dfrac{5}{x}$

$y = x + 7 + 5x^{-1}$

$\dfrac{dy}{dx} = 1 - \dfrac{5}{x^{2}}$

You try

Differentiate: $y = \dfrac{x^{2} - 4}{x - 2}$

Top is a difference of two squares — factorise it.

$y = \dfrac{(x - 2)(x + 2)}{x - 2}$

$y = x + 2$

$\dfrac{dy}{dx} = 1$

$1$

You try

Differentiate: $y = \dfrac{2x^{3} + 5x^{2} - x}{x}$

Split each term over the $x$.

$y = \dfrac{2x^{3}}{x} + \dfrac{5x^{2}}{x} - \dfrac{x}{x}$

$y = 2x^{2} + 5x - 1$

$\dfrac{dy}{dx} = 4x + 5$

$4x + 5$

You try

Differentiate: $y = \dfrac{4x^{2} - 9}{2x - 3}$

$4x^{2} - 9 = (2x - 3)(2x + 3)$ — difference of two squares.

$y = \dfrac{(2x - 3)(2x + 3)}{2x - 3}$

$y = 2x + 3$

$\dfrac{dy}{dx} = 2$

$2$

Section 13 of 13

The $f'(x)$ Notation

Same idea, different symbol. If $y = f(x)$, then the derivative is written either way:

Two notations for the same thing

$\dfrac{dy}{dx} = f'(x)$

Read $f'(x)$ as "$f$-prime of $x$" — the derivative function.

Everything you've learnt still applies — same rules, just dressed in $f$ clothes.

(i) $f(x) = 5x^{4} + 2x$

$f'(x) = 20x^{3} + 2$

(ii) $f(x) = (x + 1)(x - 3)$

$f(x) = x^{2} - 3x + x - 3$

$f(x) = x^{2} - 2x - 3$

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

(iii) $f(x) = \dfrac{1}{x^{2}}$

$f(x) = x^{-2}$

$f'(x) = -2x^{-3}$

$f'(x) = -\dfrac{2}{x^{3}}$

You try

Find $f'(x)$ when $f(x) = 7x^{3} - \sqrt{x}$.

Write $\sqrt{x}$ as $x^{\,\frac{1}{2}}$ first.

$f(x) = 7x^{3} - x^{\,\frac{1}{2}}$

$f'(x) = 21x^{2} - \dfrac{1}{2}x^{-\frac{1}{2}}$

$f'(x) = 21x^{2} - \dfrac{1}{2\sqrt{x}}$

$21x^{2} - \dfrac{1}{2\sqrt{x}}$

In the next lesson we meet the Product Rule and the Quotient Rule — for when you can't simplify into standard form. Trig, log, and exponential derivatives come with them.

That's Differentiation — Part 1.

First principles · the Rule · standard form · $f'(x)$. Coming next: Product and Quotient Rules, then Chain Rule, then logs and exponentials.