Differentiation

Question 1

Basic differentiation

Answer 1

y = axⁿ
y’ = n x axⁿ⁻¹

Question 2

Differentiation of the exponential function

Answer 2

The exponential function is the function that has itself as the derived function

This means that if f(x) = eˣ then f’(x) = eˣ

This means if f(x) = eᵍ⁽ˣ⁾ then
f’(x) = g’(x) x eᵍ⁽ˣ⁾

Question 3

Chain rule for differentiation of the exponential function

Answer 3

Chain rule:
y(u(x))
y’ = dy/dx = dy/dx . du/dx

Question 4

Tangent to curves

Answer 4

The gradient at a point on a curve is the same as the gradient of the tangent at that point

Step:
1. Find the gradient function f(x) → f’(x)
2. Substitute x₁ into the derived function to find the tangent gradient f’(x₁) = m
3. Substitute the gradient m and the point (x₁, y₁) into the equation of the tangent line

y – y₁ = m (x – x₁)
x₁ = a
y₁ = b
y – b = m(x – a)

Question 5

Differentiation of log function

Answer 5

If y = 𝑙𝑛|𝑢(𝑥)| then 𝑦′ = 𝑢′(𝑥) ×[1/𝑢(𝑥)]
, 𝑢(𝑥) > 0

The absolute value sign ensures that the argument of the log function is positive

Question 6

Using log properties

Answer 6

Common log rules can be used to help differentiate

Question 7

Normal to curves

Answer 7

The grad of the normal is the negative reciprocal

Steps:
1. Find the gradient function f(x) → f’(x)
2. Substitute x₁ into the derived function to find the tangent gradient f’(x₁) = m
3. Substitute the gradient of tangent m into m of normal
4. Sub the new gradient and the point (x₁, y₁) into the equation of the tangent line

y – y₁ = m (x – x₁)
x₁ = a
y₁ = b
y – b = m(x – a)

Question 8

Finding stationary points

Answer 8

1. Differentiate the function to find f(x)
2. Place f’(x) = 0 and solve for x
3. Sub the value into the original function to find the y-value of the stationary point
4. Determine nature of point by taking the second derivative
   y’’ > 0 (min)
   y’’ < 0 (max)

Question 9

Differentiation of exponential function with log

Answer 9

If f(x) = bˣ then f’(x) = ln(b) x bˣ

Question 10

Differentiating trigonometric functions

Answer 10

y = sin(x)
y’(x) = cos(x)

y = cos(x)
y’(x) = –sin(x)

y = tan(x)
y’(x) = sec²(x)

y = cosec(x)
y’(x) = –cosec(x) cot(x)

y = sec(x)
y’(x) = sec(x) tan(x)

y = cot(x)
y’(x) = –cosec²(x)

Question 11

Power rule

Answer 11

f(x) = [u(x)]ⁿ
f’(x) = n . [u(x)]ⁿ⁻¹ . du/dx

Question 12

Differentiating products

Answer 12

y = uv

When 2 separate functions with x in them are multiplying each other, we use the product rule. One product is u and the other is v

y = u’v + v’u

Question 13

Differentiating quotients

Answer 13

y = u / v

To use the quotient rule we need one function divided by another, the one at the top is u and the bottom one is v

y = [u’v - v’u] / v²

Question 14

Parametric equations

Answer 14

x = p + r cos(θ)
y = q + r sin(θ)

Question 15

Implicit differentiation

Answer 15

Steps
To find 𝑑𝑦/𝑑𝑥 when 𝑥 and 𝑦 are related implicitly:

a) Assume that y is a differentiable function of x
b) Differentiate both sides of the relationship with respect to x, if part of the equation consists of an expression containing y then differentiate with part with respect to to y, but immediately multiply by the correcting factor dy/dx
c) solve the resulting equation for dy/dx and make that the subject

Question 16

Derivative of tan⁻¹(x)

Answer 16

If f(x) = tan⁻¹(x)
f’(x) = 1 / x² + 1