Successive Differentiation

Last Updated : 12 Sep, 2025

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 \frac{dy}{dx}, f'(x), Dy, y', or y1. This is called the first derivative of y.
  • The derivative of \frac{dy}{dx} is denoted as \frac{d^2y}{dx^2}, f''(x), D2y, y'', or y2. This is called the second derivative of y.
  • The derivative of \frac{d^2y}{dx^2} is denoted as \frac{d^3y}{dx^3}, f'''(x), D3y, y''', or y3. This is called the third derivative of y.

Likewise, the nth derivative of y is denoted as \frac{d^ny}{dx^n}, f(n) (x), Dny, y(n)or yn.

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

  • 1st derivative: \frac{dy}{dx} =\frac{d}{dx}(x3) = 2x2
  • 2nd derivative: \frac{d^2y}{dx^2} = \frac{d}{dx}(2x2) = 2. 2 x = 4x
  • 3rd derivative: \frac{d^3y}{dx^3} = \frac{d}{dx}(4x) = 4

nth Derivatives for Some Standard Functions

Here we will discuss nth derivatives of some standard functions such as:

  • nth derivative of eax
  • nth derivative of xm
  • nth derivative of (ax + b)m
  • nth derivative of log(ax + b)
  • nth derivative of sin(ax + b)

1) nth derivative of y = eax

y1 = eax, y2 = a2eax, y3 = a3eax..

yn = aneax --(1)

Differentiating both sides w.r.t x we get

  • yn+1 = mn(memx)
  • yn+1 = mn+1emx

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

yn = aneax

2) nth derivative of y = xm

y1 = mxm-1, y2 = m(m-1)xm-2 , y3 = m(m-1)(m-2)xm-3.

yn = m(m-1)(m-2)...(m-(n-1))xm-(n+1)

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

yn = \frac{m(m-1)(m-2)(m-3)...(m-(n-1))(m-n)!}{(m-n)!}x^{m-n} .--(1

Differentiating both sides w.r.t x we get

yn+1 = \frac{m(m-1)(m-2)(m-3)...(m-(n-1))(m-n)!}{(m-(n+1))!}x^{m-(n+1)}

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

yn = \frac{m!}{(m-n)!}x^{m-n}

3) nth 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

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

Differentiating both sides w.r.t x we get

yn+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:

yn = \frac{m(m-1)(m-2)(m-3)...(m-(n-1))(m-n)!}{(m-n)!}a^n(ax+b)^{m-n} .--(1)

yn = \frac{m!}{(m-n)!}a^n(ax+b)^{m-n}

4) nth derivative of y = log(ax + b)

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

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

Differentiating both sides w.r.t x we get

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

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

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

yn = (-1)n-1an((n-1)! / (ax+b)n

5) nth 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)

yn = ansin(ax + b + nπ/2) --(1)

Differentiating both sides w.r.t x we get

yn+1 = an+!sin(ax + b + (n+1)π/2)

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

yn = ansin(ax + b + nπ/2)

Summary Table for nth Derivatives

Given below is the summary of the nth derivatives of the functions menioned above:

Function

1st derivative

nth derivative

eax

y1 = eax

yn = aneax

xm

y1 = mxm-1

yn = \frac{m!}{(m-n)!}x^{m-n}

(ax + b)m

y1 = ma(ax+b)m-1

yn = \frac{m!}{(m-n)!}a^n(ax+b)^{m-n}

log(ax + b)

y1 = a(ax+b)-1

yn = (-1)n-1an((n-1)! / (ax+b)n

sin(ax + b)

y1 = a.cos(ax + b + π/2)

yn = a.cos(ax + b + nπ/2)

cos(ax+b)

y1 = a.cos(ax + b + π/2)

yn = an.cos(ax + b + nπ/2)

eaxsin(bx + c)

y1 = reaxsin(bx + c + 𝛉)

yn = rneaxsin(bx + c + n𝛉)
a = rcos 𝛉, b = rsin 𝛉, r = √(a2 + b2)

eaxcos(bx + c)

y1 = reaxcos(bx + c + 𝛉)

yn = rneaxcos(bx + c + n𝛉)
a = rcos 𝛉, b = rsin 𝛉, r = √(a2 + b2)

1/(ax + b)

y1 = (-1)a(ax+b)-2

yn = (-1)nn!an(ax+b)-(n+1)

1/(ax+b)2

y1 = (-2)a(ax+b)-3

y2 = (-1)n(n+1)!an(ax+b)-(n+2)

Practice Questions for Successive Differentiation

1. y = xmeax, 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).xm using Leibniz Rule.

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