The derivative.

Step 1 of 4

The tangent

In the last Lesson you brought the secant’s two points together, and its slope settled on one number. The line the secant becomes as the points merge is the tangent: it touches the curve at one point and tilts exactly as the curve does there. Its slope is the derivative of the function at that point, written $f'(x)$⁠.

The board shows $y = x^2$⁠, with the point at $x = 3$⁠. The tangent is still hidden.

Step 2 of 4

The derivative as a function

At every point the tangent has a different slope, so the derivative is itself a function: for each $x$⁠ it gives the slope there. Move Point x and note the tangent’s slope at $x = 1$⁠, $2$⁠ and $3$⁠.

Why? The secant from $x$⁠ to $x + h$⁠ on $y = x^2$⁠ rises $(x + h)^2 - x^2 = 2xh + h^2$$(x + h)^2 - x^2 =$$2xh + h^2$⁠, so its slope is $\dfrac{2xh + h^2}{h} = 2x + h$$\dfrac{2xh + h^2}{h} =$$2x + h$⁠. As $h$⁠ shrinks to zero, $2x$⁠ is left.

Step 3 of 4

Where is the tangent level?

The board shows $y = x^2 - 4x + 3$⁠. At a parabola’s vertex the curve stops going down and starts going up, and there the tangent lies level: the slope is $0$⁠. That is how the derivative finds minimum and maximum points.

Step 4 of 4

The rule

The same working goes through for any quadratic function, and gives one rule:

$f(x) = ax^2 + bx + c \quad \Rightarrow \quad f'(x) = 2ax + b$$f(x) =$$ax^2 + bx + c \quad \Rightarrow$$\quad f'(x) =$$2ax + b$⁠

Each term is differentiated on its own: $ax^2$⁠ gives $2ax$⁠, $bx$⁠ gives $b$⁠ (a line, whose slope is $b$⁠), and the number $c$⁠ gives $0$⁠, because it only lifts the curve and does not change how steep it is. The board shows $y = 3x^2 - 2x + 5$⁠.