Differentiation Flashcards
Basic differentiation
y = axⁿ
y’ = n x axⁿ⁻¹
Differentiation of the exponential 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⁾
Chain rule for differentiation of the exponential function
Chain rule:
y(u(x))
y’ = dy/dx = dy/du . du/dx
Tangent to curves
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)
Differentiation of log function
If y = 𝑙𝑛|𝑢(𝑥)| then 𝑦′ = 𝑢′(𝑥) ×[1/𝑢(𝑥)]
, 𝑢(𝑥) > 0
Normal to curves
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)
Finding stationary points
Differentiate the function to find f(x)
Place f’(x) = 0 and solve for x
Sub the value into the original function to find the y-value of the stationary point
Determine nature of point by taking the second derivative
y’’ > 0 (min)
y’’ < 0 (max)
Differentiation of exponential function with log
If f(x) = bˣ then f’(x) = ln(b) x bˣ
Power rule
f(x) = [u(x)]ⁿ
f’(x) = n . [u(x)]ⁿ⁻¹ . du/dx
Differentiating products
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
Differentiating quotients
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
Parametric equations
x = p + r cos(θ)
y = q + r sin(θ)
Implicit differentiation
any y value must have dy/dx multiplied by it once differentiated
REMEMBER PRODUCT RULE FOR COMBINED TERMS