Skip to content
Related Articles

Related Articles

Improve Article
Higher Order Derivatives
  • Last Updated : 08 May, 2021

Derivative of a function f(x) tells us how will the value of the function change when we change x. This quantity gives us an idea and the direction about the rate of change of the function. For example, a positive derivative indicates the increase in the value of the function while a negative value indicates that there might be a decrease in the value of the function. Derivatives are very essential for us in predicting the limits, direction of change, and system behavior given some input. 

Derivatives and Higher-Order Derivatives

Derivative of a real function tells us about the rate of change of the function. Derivatives are defined using limits and for the function f(x), it’s derivative is denoted by f'(x). It’s definition in terms of limit is given below, 

f'(x) = \lim _{h \to 0} \frac{f(x +h) - f(x)}{h}

To calculate derivatives for different functions, we usually use the following two properties:

Multiplication Rule for Differentiation



Let’s say we have a complicated function f(x) which is multiple of two simpler functions h(x) and g(x). In that case, we use the multiplication formula for derivatives. 

\frac{d(g(x)h(x))}{dx} = g(x) \frac{d(h(x))}{dx} + h(x)\frac{g(f(x))}{dx}

Division Rule for Differentiation

In another case, let’s say our complicated function f(x) is composed to division of two different functions. For example, f(x) = \frac{g(x)}{h(x)}

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

Second Order Derivatives

Just like the derivatives tell us the rate of change of the functions, higher-order derivatives tell us the rate of change of the previous derivative. For example, a second-order derivative tells us about the rate of change of derivative. 

Let’s say we have a function f(x). 

y = f(x)



\frac{dy}{dx} = f'(x)

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

\frac{d(f'(x))}{dx} = \frac{d^2y}{dx^2}

It is also written as, f”(x). 

Let’s see some problems with second-order derivatives. 

Sample Problems

Question 1: Given f(x) = x3. Find the value of f”(x). 

Solution: 

We need to first find the derivative,

 f(x) = x3

⇒f'(x) = 3x2



Differentiating it again, we get the second order derivative. 

f”(x) = 6x

Question 2: Given f(x) = ex + sin(x). Find the value of f”(x). 

Solution: 

f(x) = ex + sin(x) 

The first derivative will be, 

f'(x) = ex + cos(x) 

Differentiating it again, 

f”(x) = ex – sin(x) 

Question 3: Given f(x) = ex.sin(x). Find the value of f”(x) at x = 0. 

Solution:  

f(x) = ex.sin(x)

Since this is product of two functions, we will use multiplication property for derivatives. 

\frac{d(f(x)g(x))}{dx} = f(x) \frac{d(g(x))}{dx} + g(x)\frac{d(f(x))}{dx}

f'(x) = exsin(x) + excos(x) 

⇒ f'(x) = ex (sin(x) + cos(x))

f”(x) = ex (sin(x) + cos(x)) + ex (cos(x) -sin(x))

⇒f”(x) = ex (2cos(x)) 

⇒f”(x) = 2excos(x) 

at x =0. 

f”(0) = 2

Question 4: Given f(x) = ex.sin(x). Find the value of f”(x) at x = 0. 

Solution: 

f(x) = ex.sin(x)

Since this is product of two functions, we will use multiplication property for derivatives. 

\frac{d(f(x)g(x))}{dx} = f(x) \frac{d(g(x))}{dx} + g(x)\frac{d(f(x))}{dx}

f'(x) = exsin(x) + excos(x) 

⇒ f'(x) = ex (sin(x) + cos(x))

f”(x) = ex (sin(x) + cos(x)) + ex (cos(x) -sin(x))

⇒f”(x) = ex (2cos(x)) 

⇒f”(x) = 2excos(x) 

Question 5: Given y = 3e2x + 2e3x, prove that \frac{d^2y}{dx^2} -5\frac{dy}{dx} + 6y =0

Solution: 

y = 3e2x + 2e3x

y’ =  6e2x + 6e3x

y”  =  12e2x + 18e3x

Substituting these values in the equation, 

\frac{d^2y}{dx^2} -5\frac{dy}{dx} + 6y =0

⇒12e2x + 18e3x – 5(6e2x + 6e3x) + 6(3e2x + 2e3x) = 0

⇒12e2x + 18e3x – 30e2x – 30e3x + 18e2x + 12e3x = 0

⇒-30e2x + 30e3x – 30e2x – 30e3x = 0

⇒ 0 = 0 

Hence, Proved. 

Question 6: Given y = ex(x + 1). Find the value of second derivative at x = 1. 

Solution: 

 y = ex(x + 1)

Since this function is product of two functions, We will use multiplication rule for derivative. 

\frac{d(f(x)g(x))}{dx} = f(x) \frac{d(g(x))}{dx} + g(x)\frac{d(f(x))}{dx}

y’ = ex (x + 1) + ex

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

y”= \frac{d(e^x(x+1))}{dx} + \frac{d(e^x)}{dx}

Again this function will require multiplication rule for differentiation. 

y” = ex (x + 1) + ex + ex

⇒ y” = ex(x + 3) 

Question 7: Given y = \frac{x}{x^2 + 1}. Find the value of second derivative at x = 1. 

Solution: 

 y =\frac{x}{x^2 + 1}

Since this function is division of two functions, We will use division rule for derivative. 

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

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. 

Again this function will require multiplication rule for differentiation. 

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

My Personal Notes arrow_drop_up
Recommended Articles
Page :