The first derivative, f'(x), tells us critical information about the behavior of the original function f(x).
As seen in Unit 1, f'(a) is the slope of the tangent line to the curve y = f(x) at the point x = a.
The sign of f'(x) determines if the function f(x) is increasing or decreasing on an interval.
Points where f'(x) = 0 or f'(x) is undefined are called critical points. These are potential locations for local maxima or minima.
The derivative dy/dx represents the instantaneous rate of change of y with respect to x.
This is a common application. If two or more quantities are related by an equation and are all changing over time (t), we can find the relationship between their rates of change by differentiating the equation with respect to time.
Successive differentiation means differentiating a function repeatedly.
The second derivative, f''(x), has its own significance:
Finding a general formula for the n-th derivative is a key skill. This is usually done by calculating the first few derivatives (f', f'', f''') and looking for a pattern.
| Function, y = f(x) | n-th Derivative, yⁿ |
|---|---|
| eᵃˣ | aⁿ * eᵃˣ |
| aˣ | (ln a)ⁿ * aˣ |
| xᵐ | [m(m-1)...(m-n+1)] * xᵐ⁻ⁿ = [m! / (m-n)!] * xᵐ⁻ⁿ |
| (ax + b)ᵐ | aⁿ * [m! / (m-n)!] * (ax + b)ᵐ⁻ⁿ |
| 1 / (ax + b) | (-1)ⁿ * n! * aⁿ * (ax + b)⁻(ⁿ⁺¹) |
| ln(ax + b) | (-1)ⁿ⁻¹ * (n-1)! * aⁿ * (ax + b)⁻ⁿ |
| sin(ax + b) | aⁿ * sin(ax + b + nπ/2) |
| cos(ax + b) | aⁿ * cos(ax + b + nπ/2) |
| eᵃˣ * sin(bx + c) | rⁿ * eᵃˣ * sin(bx + c + nθ), where a = r*cos(θ), b = r*sin(θ) |
| eᵃˣ * cos(bx + c) | rⁿ * eᵃˣ * cos(bx + c + nθ), where a = r*cos(θ), b = r*sin(θ) |
To find the n-th derivative of a rational function (a polynomial divided by another polynomial), the key is to first decompose it using partial fractions.
Example: Find the n-th derivative of y = 1 / (x² - a²).
Solving this gives A = 1/(2a) and B = -1/(2a).
y = [1/(2a)] * [1/(x-a)] - [1/(2a)] * [1/(x+a)]We use the formula: dⁿ/dxⁿ [1/(ax+b)] = (-1)ⁿ * n! * aⁿ * (ax+b)⁻(ⁿ⁺¹)
yⁿ = [1/(2a)] * [(-1)ⁿ n! (1)ⁿ (x-a)⁻(ⁿ⁺¹)] - [1/(2a)] * [(-1)ⁿ n! (1)ⁿ (x+a)⁻(ⁿ⁺¹)]This theorem provides a formula for the n-th derivative of a product of two functions.
If y = u * v, where u and v are functions of x, then the n-th derivative of y is: (uv)ⁿ = ⁿC₀ uⁿv + ⁿC₁ uⁿ⁻¹v¹ + ⁿC₂ uⁿ⁻²v² + ... + ⁿCᵣ uⁿ⁻ʳvʳ + ... + ⁿCₙ uvⁿ Where:
- uⁿ is the n-th derivative of u.
- vʳ is the r-th derivative of v.
- ⁿCᵣ = n! / [r! * (n-r)!] is the binomial coefficient.
Common Problem: Find the n-th derivative of y = x² * eᵃˣ. Here, let u = eᵃˣ (its n-th derivative is known) and v = x² (its derivatives become zero).
v = x², v¹ = 2x, v² = 2, v³ = 0 (and all higher derivatives are 0).
u = eᵃˣ, uⁿ = aⁿeᵃˣ, uⁿ⁻¹ = aⁿ⁻¹eᵃˣ, uⁿ⁻² = aⁿ⁻²eᵃˣ
yⁿ = (ⁿC₀ uⁿv) + (ⁿC₁ uⁿ⁻¹v¹) + (ⁿC₂ uⁿ⁻²v²) + (rest are zero)
yⁿ = (1 * aⁿeᵃˣ * x²) + (n * aⁿ⁻¹eᵃˣ * 2x) + ([n(n-1)/2] * aⁿ⁻²eᵃˣ * 2)
yⁿ = aⁿ⁻²eᵃˣ * [a²x² + 2nax + n(n-1)]
When evaluating a limit, we sometimes get an undefined expression. These are called indeterminate forms. L'Hospital's Rule is a tool to solve limits that result in these forms.
The main indeterminate forms are:
Other forms can be converted into these two:
For the exponential forms (1∞, 0⁰, ∞⁰), let the limit be L. Take the natural logarithm (ln L) on both sides. This will turn the limit into a 0 * ∞ form, which can then be converted to 0/0 or ∞/∞.
L'Hospital's Rule is a method for evaluating limits that result in the indeterminate forms 0/0 or ∞/∞.
Suppose lim (x → a) f(x) = 0 and lim (x → a) g(x) = 0.
OR
Suppose lim (x → a) f(x) = ±∞ and lim (x → a) g(x) = ±∞.
Then: lim (x → a) [f(x) / g(x)] = lim (x → a) [f'(x) / g'(x)] ...provided the limit on the right side exists (or is ±∞).
Evaluate L = lim (x → 0) (1 + x)¹/ˣ