Successive Differentiation

Successive differentiation is the process of differentiating a function repeatedly, as many times as required. The derivatives obtained at each stage are called successive derivatives or higher-order derivatives.

Uses of Successive Derivatives

• Expanding functions into series (e.g., Taylor or Maclaurin series)
• Solving and forming differential equations
• Studying the shape and behaviour of curves (e.g., slope, concavity, acceleration)

Notation

If f(x) is a function defined on a domain D:

• The derivative of y with respect to x is written as , f'(x), Dy, y', or y₁. This is called the first derivative of y.
• The derivative of is denoted as , f''(x), D²y, y'', or y₂. This is called the second derivative of y.
• The derivative of is denoted as , f'''(x), D³y, y''', or y₃. This is called the third derivative of y.

Likewise, the n^th derivative of y is denoted as , f⁽ⁿ⁾ (x), Dⁿy, y⁽ⁿ⁾or y_n.

For example: y = x³, find the 3^rd derivative of the equation.

• 1st derivative: =(x³) = 2x²
• 2nd derivative: = (2x²) = 2. 2 x = 4x
• 3rd derivative: = (4x) = 4

n^th Derivatives for Some Standard Functions

Here we will discuss n^th derivatives of some standard functions such as:

• n^th derivative of e^ax
• n^th derivative of x^m
• n^th derivative of (ax + b)^m
• n^th derivative of log(ax + b)
• n^th derivative of sin(ax + b)

1) n^th derivative of y = e^ax

y₁ = e^ax, y₂ = a²e^ax, y₃ = a³e^ax..

y_n = aⁿe^ax --(1)

Differentiating both sides w.r.t x we get

• y_n+1 = mⁿ(me^mx)
• y_n+1 = mⁿ⁺¹e^mx

Therefore by mathematical induction we can say that (1) is true for all integer n

y_n = aⁿe^ax

2) n^th derivative of y = x^m

y₁ = mx^m-1, y₂ = m(m-1)x^m-2 , y₃ = m(m-1)(m-2)x^m-3.

y_n = m(m-1)(m-2)...(m-(n-1))x^m-(n+1)

Modifying the equation by multiplying and dividing with (m-n)!, we get:

y_{n =} .--(1

Differentiating both sides w.r.t x we get

y_n+1 =

Therefore by mathematical induction we can say that (1) is true for all integer n

y_n =

3) n^th derivative of y = (ax + b)^m

y₁ = ma(ax+b)^m-1, y₂ = m(m-1)a²(ax+b)^m-2, y₃ = m(m-1)(m-2)(a³(ax+b)^m-3

y_n =m(m-1)(m-2)....(m-(n-1))(aⁿ(ax+b)^m-n--(1)

Differentiating both sides w.r.t x we get

y_n+1 = m(m-1)(m-2)....(m-(n-1))(m-n)aⁿ⁺¹(ax+b)^m-(n+1)

Therefore by mathematical induction we can say that (1) is true for all integer n

Modifying the equation 1 by multiplying and dividing with (m-n)!, we get:

y_{n =} .--(1)

y_n =

4) n^th derivative of y = log(ax + b)

y₁ = a(ax+b)^-1, y₂ = (-1)a²(ax+b)^-2, y₃ = (-1)(-2)(a³(ax+b)^-3

y_n =(-1)(-2)....(-(n-1))(aⁿ(ax+b)^-n--(1)

Differentiating both sides w.r.t x we get

y_n+1 = (-1)(-2)....(-(n-1))(-n)aⁿ⁺¹(ax+b)^-(n+1)

y_n and y_n+1 is in the same form, therefore by mathematical induction we can say that (1) is true for all integers n.

y_n = (-1)(-2)....(-(n-1))(aⁿ(ax+b)^-n

y_n = (-1)^n-1aⁿ((n-1)! / (ax+b)ⁿ

5) n^th derivative of y = sin(ax + b)

y₁ = a.cos(ax + b) = a.cos(ax + b + π/2)

y₂ = a²cos(ax + b + π/2) = a²sin(ax + b + 2π/2) ,

y₃ =a³cos(ax + b + 2π/2) = a³sin(ax + b + 3π/2)

y_n = aⁿsin(ax + b + nπ/2) --(1)

Differentiating both sides w.r.t x we get

y_n+1 = a^n+!sin(ax + b + (n+1)π/2)

y_n and y_n+1 is in the same form, therefore by mathematical induction we can say that (1) is true for all integers n.

y_n = aⁿsin(ax + b + nπ/2)

Summary Table for n^th Derivatives

Given below is the summary of the n^th derivatives of the functions menioned above:

Practice Questions for Successive Differentiation

1. y = x^me^ax, find the nth derivative.

2. y = (1−x)^−m (integer m≥1), find the nth derivative

3. y = 1/(x² + a²), find the nth derivative.

4. Find the derivative of y = sin(ax).cos(ax) using Leibniz Rule.

5. Find the derivative of y = log(x).x^m using Leibniz Rule.