Rules Of Differentiation SS2 Further Mathematics Lesson Note

DIFFERENTIATION OF A SUM

If y = u(x) + v(x), where u and v are functions of x, then

dy/dx = du/dx + dv/dx

Example 1:

Find the derivative of y = 3x⁵ – 2x⁴ – 2x²

Solution

y = 3x⁵ – 2x⁴ – 2x²

dy/dx = 5 × 3x⁵⁻¹ – 4 × 2x⁴⁻¹ – 2 × 2x²⁻¹

dy/dx = 15x⁴ – 8x³ – 4x

Example 2:

Find the derivative of y = 3x⁵ – 2x⁴ + 6x³ + 3x – 11

Solution

y = 3x⁵ – 2x⁴ + 6x³ + 3x – 11

dy/dx = 5 × 3x⁵⁻¹ – 4 × 2x⁴⁻¹ + 3 × 6x³⁻¹ + 1 × 3x¹⁻¹ – 0 × 11x⁰

dy/dx = 15x⁴ – 8x³ + 18x² + 3

DIFFERENTIATING OF TRIGONOMETRIC (TRANSCENDENTAL) FUNCTIONS

I. If y = sin x then dy/dx = cos x

II. If y = cos x then dy/dx = –sin x

III. If y = tan x then dy/dx = sec²x

IV. If y = cot x then dy/dx = –cosec²x

V. If y = sec x then
  dy/dx = sec x tan x

VI. If y = logₑ x or ln x then dy/dx = 1/x

VII. If y = eˣ then dy/dx = eˣ

If the angle is multiplied by a constant, for example if x is a constant, then:

I. If y = sin kx then dy/dx = k cos kx

II. If y = cos kx then dy/dx = –k sin kx

III. If y = tan kx then dy/dx = k sec²kx

IV. If y = cot kx then dy/dx = –k cosec²kx

V. If y = sec kx then dy/dx = k sec kx tan kx

Example: Differentiate y = 3e⁵ˣ

Solution

  y = 3e⁵ˣ
  dy/dx = 3 × 5e⁵ˣ
  dy/dx = 15e⁵ˣ

FUNCTION OF FUNCTION (CHAIN RULE)

Suppose that y is a function of u and u is a function of x. In other words, if y = f(u) and u = g(x), then:

  dy/dx = (dy/du) × (du/dx)

The above illustration is called the chain rule for differentiation.

Example 1:

Find the derivative of y = (x⁷ – 12)²

Solution

Let u = x⁷ – 12, then y = u²

So, dy/du = 2u and du/dx = 7x⁶

Hence, dy/dx = (dy/du) × (du/dx)

     = 2(x⁷ – 12) × 7x⁶
     = 14x⁶(x⁷ – 12)

Example 2:

Find the derivative of y = sin 6x

Solution

Let u = 6x, then y = sin u

Therefore, dy/du = cos u and du/dx = 6

Hence, dy/dx = (dy/du) × (du/dx)

     = cos u × 6
     = 6 cos 6x

PRODUCT RULE

If y = u v, where u and v are functions of x, then

dy/dx = u(dv/dx) + v(du/dx)

Example: Find the derivative of y = x^2 sin 6x.

Solution:

y = x^2 sin 6x

Let u = x^2 and v = sin 6x. Then

du/dx = 2x
dv/dx = 6 cos 6x

Therefore,

dy/dx = u(dv/dx) + v(du/dx)

= x^2(6 cos 6x) + sin 6x(2x)

= 6x^2 cos 6x + 2x sin 6x