10A

Tangents and the derivative

The derivative as the gradient of the tangent, sketching the gradient function, derivatives of linear and constant functions and semicircles, and secants versus tangents.

Tangents and the derivative

A tangent touches a curve at a point and has the same steepness as the curve there. A secant (or chord) joins two points on the curve.

The tangent and the gradient function

Orange: y = f(x). Blue dashed: the tangent at x = a. Faint green: the gradient function y = f′(x). Its height at a is the gradient of the tangent.

Try this

Find the two places where the tangent is horizontal. Where does the green curve cross the $x$-axis?
Where the orange curve goes down, what sign does the green curve have?

Sketching the gradient function

Where the curve is going up, $f'(x) > 0$: the derivative graph is above the axis.
Where the curve is going down, $f'(x) < 0$: the derivative graph is below the axis.
Where the tangent is horizontal, $f'(x) = 0$: the derivative graph meets the axis.

Lines and semicircles

Example 1: a semicircle●●

Find the gradient of the tangent to $y = \sqrt{25 - x^2}$ at $(3, 4)$.

The radius from $(0, 0)$ to $(3, 4)$ has gradient $\tfrac43$.

A tangent to a circle is perpendicular to the radius.

The tangent has gradient $-\tfrac34$, so $f'(3) = -\tfrac34$.

Flashcards

$f'(a)$ means

The gradient of the tangent at $x = a$.

Derivative of $mx + b$

$m$

Derivative of a constant

$0$

Curve going down

$f'(x) < 0$

Practice

$f'(a)$ is:
If $f(x) = 5x - 2$, find $f'(x)$.
If $f(x) = 7$, find $f'(x)$.
The gradient of the tangent to $y = \sqrt{25 - x^2}$ at $(3, 4)$ is:
The tangents to $y = x^2$ at $x = -1, 0, 1, 2$ have gradients $-2, 0, 2, 4$. The gradient function is:
Where a curve is going down, the derivative is:
A concave up parabola has its vertex at $x = 1$. Its derivative graph is:
Find the gradient of the secant on $y = x^2$ from $x = 1$ to $x = 3$.
The tangent to $y = f(x)$ at a point where $f'(x) = 0$ is: