Summary — What This Chapter Covers
- dy/dx is the limit of the gradient of a chord as the chord shrinks to a point.
- d/dx(xn) = nxn−1 for any rational n.
- Chain rule: dy/dx = dy/du × du/dx, so d/dx(ax + b)n = an(ax + b)n−1.
- Tangent gradient = dy/dx; normal gradient = −1/(dy/dx).
- f is increasing where f′(x) > 0 and decreasing where f′(x) < 0.
- Connected rates of change: dA/dt = dA/dr × dr/dt.
- Stationary points where dy/dx = 0; d2y/dx2 > 0 minimum, < 0 maximum.
1. Derivatives and the Chain Rule
Derivatives
From chord gradients to nxn−1
The gradient of the chord joining the points with x-coordinates 2 and 2 + h on y = x3 is [(2 + h)3 − 8]/h = 12 + 6h + h2. As h → 0 this tends to 12, which is the gradient of the curve at x = 2. You need only this informal idea of a limit; then use the rules.
Before differentiating, rewrite roots and fractions as powers: √x = x½, 2/x = 2x−1. For a function of a function, use the chain rule.
Rules
d/dx(xn) = nxn−1 · dy/dx = (dy/du)(du/dx) · notation f′(x), f″(x), dy/dx, d2y/dx2
Worked Example
Find dy/dx when y = √(2x3 + 5).
Practice Question 1
Differentiate y = 6√x − 2/x.
Practice Question 2
Differentiate y = 4/(3x − 1)2.
2. Tangents, Normals and Rates of Change
Applications
Using the gradient
At the point (x1, y1), find m = dy/dx there; the tangent is y − y1 = m(x − x1) and the normal has gradient −1/m. For increasing or decreasing functions, solve the inequality f′(x) > 0 or f′(x) < 0.
For connected rates of change, link the rate you want to the rate you know with the chain rule, then substitute the given value at the end.
Connected rates
dA/dt = (dA/dr) × (dr/dt)
Worked Example
Find the equations of the tangent and normal to y = x3 − 3x2 + 2 at the point where x = 3.
Worked Example
The radius of a circle increases at 0.5 cm s−1. Find the rate of increase of the area when r = 4 cm.
Practice Question
The volume of a sphere increases at 20 cm3 s−1. Find the rate of increase of the radius when r = 5 cm. (V = 4/3 πr3.)
3. Stationary Points
Stationary points
Where dy/dx = 0
2
2
Nature
d2y/dx2 > 0 ⇒ minimum · d2y/dx2 < 0 ⇒ maximum
Worked Example
Find the stationary points of y = 2x3 − 3x2 − 12x + 5 and determine their nature.
Memorise