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Differentiation

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  Cambridge International AS & A Level Mathematics 9709 — Pure Mathematics 1

Differentiation

The big idea: The derivative is the gradient of the curve, found as the limit of chord gradients. Once you can find it, you can write tangents, measure rates of change and locate maxima and minima.

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.

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