Higher Order Derivatives

Higher order derivatives refer to the derivatives of a function that are obtained by repeatedly differentiating the original function.

• The first derivative of a function, f′(x), represents the rate of change or slope of the function at a point.
• The second derivative, f′′(x), is the derivative of the first derivative and measures the curvature or concavity of the function.
• The third derivative, f′′′(x), is the derivative of the second derivative, and so on.

In general, the n^th derivative of a function f(x) is denoted as fⁿ(x) and represents the rate of change of the rate of change... (continuing for higher-order derivatives).

Second Order Derivatives

Second Order Derivative tells us about the rate of change of the derivative of a function. It is used to find the maximum and minimum values of a function. It also gives the optimal solution of any equation including functions.  We find the second-order derivative of any function by further derivating the function.

Let's say we have a function y = f(x). 

dy/dx = f'(x)

If f'(x) is a differentiable function, we can differentiate it again to get a second-order derivative. It is denoted by, 

d/dx{f'(x)} = d²y/dx²

Second Order Derivatives are represented as, f''(x).

Example: Given y = x/(x² + 1). Find the value of the second derivative at x = 1

Solution:

Given: y = x/(x² + 1)

From quoteint rule as\frac{d}{dx}(\frac{g(x)}{h(x)}) = \frac{g(x)h'(x) - h(x)g'(x)}{h(x)^2}

Thus, y' = \frac{(x^2+1) - x(2x)}{(x^2 + 1)^2}
⇒ y' = \frac{1 - x^2}{(x^2 + 1)^2}

Now we can differentiate it again to get the second derivative.

y''=\frac{(x^2 + 1)^2(-2x) - (1 - 2x^2)(2(x^2+1)2x)}{(x^2 + 1)^4}    

At x = 1

⇒ y'' = \frac{(1^2 + 1)^2(-2) - (1 - 21^2)(2(1^2+1)2)}{(1^2 + 1)^4}
⇒ y'' = \frac{(2)^2(-2) - (1 - 2)(2(2)2)}{(2)^4}
⇒ y'' = \frac{-8 + 8}{(2)^4}
⇒ y'' = 0

Thus, second order derivative of y = x/(x² + 1) at x = 1 is, y'' = 0.

Third Order Derivative

Third Order Derivative tells us about the rate of change of derivative of second order. We find the third order derivative of any function by further derivating the second order derivative of the function iff differentiation is possible.

Let's say we have a function f(x). 

y = f(x)

dy/dx = f'(x)

If f'(x) is differentiable function, we can differentiate it again to get a second-order derivative. It is denoted by, 

f''(x) = d/dx{f'(x)} = d²y/dx²

Differentiating it again with respect to x

d/dx{f''(x)} = d³y/dx³

Third Order Derivative are represented as, f'''(x).

Example: Given y(x) = 3x³ + 12x + 4. Find the value of third derivative at x = 1

Solution:

Given,

y(x) = 3x³ + 12x + 4

Differentiating with respect to x

y'(x) = 3(3x²) + 12(1) + 0

y'(x) = 9x² + 12 {This first derivative}

Again differentiating with respect to x

y''(x) = 9(2x) + 0

y''(x) = 18x {This second derivative}

Again differentiating with respect to x

y'''(x) = 18(1) = 18 {This required third derivative}

at x = 1

y'''(1) = 18

Higher-Order Derivative in Parametric Form

We can also find higher order derivative of the function in the parametric form. Suppose we are given with functions y(t) and x(t) and they both are the function of parameter 't'. Then we can easily find the first order derivative as,

dy/dx = (dy/dt) × (dt/dx)

Now for finding the higher order derivative. Let dy/dt = y'(t) and dx/dt = x'(t) then,

d²y/dx² = d/dx(dy/dx)
= {d/dt[(dy/dt) × (dt/dx)]}/dx/dt
= d/dt[y'(t)/x'(t)] × dt/dx

Read More about Differentiation Formula.

Application of Higher Order Derivative

There are various applications of the Higher Order Derivative such as,

• To find the Acceleration of the any length-time function.
• To find maxima and minima of any function
• To find the shape of various graphs.
• For Second Derivative Test, etc.

Read More,

• Differentiation and Integration Formula
• Differential Equations
• Advanced Differentiation

Examples on Higher Order Derivative

Example 1: Given f(x) = x³. Find the value of third derivative of f(x), i.e. f'''(x). 

Solution: 

f(x) = x³
f'(x) = 3x²

Differentiating it again
f''(x) = 6x

Differentiating it again
f'''(x) = 6

Thus, third derivative of f(x) = x³ is 6.

Example 2: Given f(x) = e^x + sin(x). Find the value of f'''(x) at x = 0

Solution: 

f(x) = e^x + sin(x) 

First derivative is,
f'(x) = e^x + cos(x) 

Differentiating it again, 
f''(x) = e^x - sin(x)

Differentiating it again, 
f'''(x) = e^x - cos(x)

at x = 0
f'''(0) = e⁰ - cos (0) = 1 - 1 = 0
⇒ f'''(x) = 0

Example 3: Given f(x) = e^x.sin(x). Find the value of f''(x) at x = 0. 

Solution:  

f(x) = e^x.sin(x)

As we know, \frac{d(f(x)g(x))}{dx} = f(x) \frac{d(g(x))}{dx} + g(x)\frac{d(f(x))}{dx}

⇒ f'(x) = e^xsin(x) + e^xcos(x) 
⇒ f'(x) = e^x (sin(x) + cos(x))
⇒ f''(x) = e^x (sin(x) + cos(x)) + e^x (cos(x) -sin(x))
⇒ f''(x) = e^x (2cos(x)) 
⇒ f''(x) = 2e^xcos(x) 

at x =0
f''(0) = 2e⁰cos(0)
⇒ f''(0) = 2(1)(1)
⇒ f''(0) = 2

Example 4: Given f(x) = e^x.sin(x). Find the value of f''(x) at x = 0. 

Solution: 

f(x) = e^x.sin(x)

As we know,\frac{d(f(x)g(x))}{dx} = f(x) \frac{d(g(x))}{dx} + g(x)\frac{d(f(x))}{dx}

⇒ f'(x) = e^xsin(x) + e^xcos(x) 
⇒ f'(x) = e^x (sin(x) + cos(x))
⇒ f''(x) = e^x (sin(x) + cos(x)) + e^x (cos(x) -sin(x))
⇒ f''(x) = e^x (2cos(x)) 
⇒ f''(x) = 2e^xcos(x) 

Example 5: Given y = 3e^2x + 2e^3x, prove that \frac{d^2y}{dx^2} -5\frac{dy}{dx} + 6y =0

Solution: 

y = 3e^2x + 2e^3x
⇒ y' =  6e^2x + 6e^3x
⇒ y''  =  12e^2x + 18e^3x

Substituting these values in the equation, 

\frac{d^2y}{dx^2} -5\frac{dy}{dx} + 6y =0
⇒ 6e^2x + 18e^3x - 5(6e^2x + 6e^3x) + 6(3e^2x + 2e^3x) = 0
⇒ 12e^2x + 18e^3x - 30e^2x - 30e^3x + 18e^2x + 12e^3x = 0
⇒ 30e^2x + 30e^3x - 30e^2x - 30e^3x = 0
⇒ 0 = 0 

Hence, Proved. 

Example 6: Given y = e^x(x + 1). Find the value of second derivative at x = 1. 

Solution: 

 y = e^x(x + 1)

As we know, \frac{d(f(x)g(x))}{dx} = f(x) \frac{d(g(x))}{dx} + g(x)\frac{d(f(x))}{dx}

⇒ y' = e^x (x + 1) + e^x

Now differentiate it again,

y''= \frac{d(e^x(x+1))}{dx} + \frac{d(e^x)}{dx}
⇒ y'' = e^x (x + 1) + e^x + e^x
⇒ y'' = e^x(x + 3)