Slope & the Differentiation Rule

Section 1 of 3

Slope of a Line

Differentiation is about slope. Before the slope of a curve, be solid on the slope of a straight line: change in $y$ over change in $x$.

Slope

$m = \dfrac{\text{rise}}{\text{run}} = \dfrac{\text{change in }y}{\text{change in }x} = \dfrac{dy}{dx}$. Two points: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$. $y = mx + c \Rightarrow$ slope $m$. $ax + by + c = 0 \Rightarrow m = -\dfrac{a}{b}$.

Worked example — slope through two points

Find the slope of the line through $(-3, 0)$ and $(0, 3)$.

$(x_1,y_1) = (-3, 0)$, $(x_2,y_2) = (0, 3)$

$m = \dfrac{3 - 0}{0 - (-3)} = \dfrac{3}{3}$

$m = 1$

Worked example — already y = mx + c

Find the slope of $y = 3x + 1$.

The slope is the number in front of $x$

$m = 3$

Worked example — rearrange first

Find the slope of $5x + y = 6$.

$y = -5x + 6$

$m = -5$

You try

Find the slope of $2x + 3y = 6$.

Rearrange to $y = mx + c$, or use $m = -\tfrac{a}{b}$ with $a=2,\ b=3$.

$3y = -2x + 6$, $y = -\tfrac{2}{3}x + 2$

or $m = -\tfrac{a}{b} = -\tfrac{2}{3}$

$m = -\tfrac{2}{3}$

You try

Find the slope of $4x + 5y = 9$.

Use $m = -\tfrac{a}{b}$ with $a=4,\ b=5$.

$m = -\tfrac{a}{b} = -\tfrac{4}{5}$

$m = -\tfrac{4}{5}$

Section 2 of 3

The Differentiation Rule

A line has one slope everywhere; a curve’s slope changes point to point. Differentiation gives a formula for that changing slope.

The rule

$y = x^n \Rightarrow \dfrac{dy}{dx} = n\,x^{n-1}$ — multiply by the power, reduce the power by 1. A constant differentiates to $0$. And the headline: $\dfrac{dy}{dx} = m = $ slope.

Worked example — a square

Differentiate $y = x^2$.

Multiply by $2$, reduce the power to $1$

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

Worked example — term by term

Differentiate $y = 3x^2 + 7x + 6$.

The constant $6$ vanishes

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

Worked example — a cubic

Differentiate $y = x^3 + 7x^2 + 8x + 3$.

$x^3 \to 3x^2$, $7x^2 \to 14x$, $8x \to 8$, constant $\to 0$

$\dfrac{dy}{dx} = 3x^2 + 14x + 8$

You try

Differentiate $y = 8 - 11x - 3x^2$.

Mind the signs. Constant $\to 0$, $-11x \to -11$, $-3x^2 \to -6x$.

$\dfrac{dy}{dx} = -11 - 6x$

You try

Differentiate $y = 6x^3 + 8x^2 + 11x + 20$.

$x^3$ term times $3$, $x^2$ term times $2$, $x$ term keeps its number, constant gone.

$\dfrac{dy}{dx} = 18x^2 + 16x + 11$

You try

Differentiate $y = 7 - 3x - 2x^2$.

Constant $\to 0$, $-3x \to -3$, $-2x^2 \to -4x$.

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

Section 3 of 3

Slope at a Point

Once you have $\dfrac{dy}{dx}$, you have a formula for the slope. For the slope at a particular $x$-value, sub that value in. Differentiate, then substitute.

Worked example — slope at one point

$y = 1 + 3x - 6x^2$. Find the slope when $x = 1$.

$\dfrac{dy}{dx} = 3 - 12x$

At $x = 1$: $m = 3 - 12(1) = -9$

$m = -9$

Worked example — slope at two points

$y = x^2 - 3x + 5$. Find the slope when $x = -5$ and when $x = 5$.

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

At $x = -5$: $m = 2(-5) - 3 = -13$

At $x = 5$: $m = 2(5) - 3 = 7$

$m = -13$ or $m = 7$

You try

$y = 3x^2 - 6x + 9$. Find the slope when $x = -7$ and when $x = 9$.

Differentiate to $6x - 6$, then sub each $x$ in.

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

At $x = -7$: $m = 6(-7) - 6 = -48$

At $x = 9$: $m = 6(9) - 6 = 48$

$m = -48$ or $m = 48$

You try

$y = 5x^2 - 9x + 3$. Find the slope when $x = -2$ and when $x = 8$.

Differentiate to $10x - 9$, then sub in.

$\dfrac{dy}{dx} = 10x - 9$

At $x = -2$: $m = 10(-2) - 9 = -29$

At $x = 8$: $m = 10(8) - 9 = 71$

$m = -29$ or $m = 71$

That’s Class 1.

Slope of a line, the differentiation rule, and the slope at a point. Class 2: the tangent to a curve, and finding points & unknowns.