What are Higher Order Derivatives?

Do you know that we can find the multiple derivatives of the same function?  Derivatives beyond the first derivative are known as higher order derivatives. In this article, we will explain how to find the higher order derivatives of a function with some examples. First, let us see what are the higher order derivatives of the function and how we can calculate them.

If the derivative of a function, the first derivative, f'(x), is differentiated, a new function is obtained called the second derivative, f''(x).

If this function is differentiated again, a third derivative, f'''(x), is obtained.

If the third derivative, f'''(x), is differentiated, the fourth derivative, f'v(x), is obtained. This process can continue, and these resultant functions are referred to as higher order derivatives.

Let us now explain higher order derivatives in this way. Suppose there is a function y = f(x). If this function has a finite derivative f'(x) in a specific interval, then the derivative of the function denoted by f ' (x) is also a function in this interval. We can find the second derivative of a function that is differentiable. Different notations of the second derivatives are given below:

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

Now, if the second derivative of the function is differentiable further, then we can find the third derivative of the function. The notation for the third derivative of the function is given below:

y ''' = f ''' = \frac {d ^3 y} {d x ^3}

Different notations. mean that the second and third derivatives can be denoted using any of the above notations. Now, let us proceed to some of the examples in which we will find higher order derivatives of the functions.

Example 1

Calculate the 1st, 2nd, 3rd and 4th derivatives of the following function:

f(x) = 2x ^3 - 15x ^2 +36x - 12

Step 1 - First Derivative

Let us compute the first derivative of the above function first. To calculate the derivative of the above function, we will use multiple derivative rules which include sum and difference, power and constant rule.

f ' (x) = 6 x ^ 2 - 30x + 36

Step 2 - Second Derivative

Now, we will compute the second derivative of the function. For this, we will simply differentiate the first derivative f ' (x) = 6 x ^ 2 - 30x + 36 of the function using various rules of derivatives like this:

f '' (x) = 12x - 30

Step - 3 Third Derivative

We will calculate the third derivative by differentiating the second derivative of the function f '' (x) = 12x - 30 further like this:

f ''' (x) = 12

Step 4 - Fourth Derivative

Now, we will find the derivative of the function obtained after calculating the third derivative f ''' (x) = 12 like this:

f ^ {iv} = 0

Because the derivative of the constant is 0, so the fourth derivative will be 0. Now, the function cannot be differentiated any further.

Example 2

Calculate the 1st, 2nd, 3rd, 4th and 5th derivatives of the following function:

f(x) = 5x ^4 + 6x ^3 - 7 x ^ 2 +3x - 7

Step 1 - First Derivative

Let us compute the first derivative of the above function first. To calculate the derivative of the above function, we will use multiple derivative rules which include sum and difference, power and constant rule.

f ' (x) = 20 x ^ 3 + 18 x ^2 - 14x + 3

Step 2 - Second Derivative

Now, we will compute the second derivative of the function. For this, we will simply differentiate the first derivative f ' (x) = 20 x ^ 3 + 18 x ^2 - 14x + 3 of the function using various rules of derivatives like this:

f '' (x) = 60 x ^ 2 + 36 x - 14

Step - 3 Third Derivative

We will calculate the third derivative by differentiating the second derivative of the function f '' (x) = 60 x ^ 2 + 36 x - 14 further like this:

f ''' (x) = 120 x + 36

Step 4 - Fourth Derivative

Now, we will find the derivative of the function obtained after calculating the third derivative f ''' (x) = 120 x + 36  like this:

f ^ {iv} = 120

Step 5 - Fifth Derivative

Now, we will find the fifth derivative of the function f ^ {iv} = 120 like this:

f ^ {v} = 0

Because the derivative of the constant is 0, so the fifth derivative of the constant 120 will be 0.  It is obvious that 0 cannot be differentiated further, so we will stop our differentiation here.

Example 3

Calculate the 1st, 2nd and 3rd derivatives of the following function:

f(x) = \sqrt {3x}

First Derivative

We can write the above function as f (x) = \sqrt {3} \cdot \sqrt {x}. We will apply the derivative root rule on \sqrt {x}. According to the derivative root rule, \sqrt {x} = \frac{1}{2 \sqrt {x}}. Hence, the first derivative of the function will be:

f ' (x) = \frac {\sqrt {3}} {2 \sqrt {x}}

Second Derivative

Now, we will find the second derivative of the above function f ' (x) = \frac {\sqrt {3}} {2 \sqrt {x}} which is obtained after differentiating the original function.

f '' (x) =  - \frac {1}{2 x ^ \frac {3} {2}}

Third Derivative

In this step, we need to calculate the third derivative of the function obtained in the last step f '' (x) =  - \frac {1}{2 x ^ \frac {3} {2}}.

f ''' (x) = - \frac {1} {2} \cdot \frac {1} { x ^ \frac {3} {2}}

Since, \frac {1} {x ^ \frac {3} {2}} can be written as x ^ \frac - {3}{2}, hence we can use the power rule here like this:

f ''' (x) = - \frac {(- \frac {3}{2}) x ^ {-\frac {3} {2} - 1}} {2}

= \frac {3} {4 x ^ {\frac {5}{2}}}

Example 4

Find the 1st, 2nd and 3rd derivatives of the following function:

f(x) = sin ^ 3 (x)

Step 1- First Derivative

To find the first derivative of the function, we will apply the derivative chain rule here. The simplest way to apply it is to use the derivative power rule and then multiply with the derivative of the inner term. Note that the derivative of sin x = cos x.

f ' (x) = 3 sin ^2 x cos x

Step 2 - Second Derivative

Now, we will compute the second derivative of the function f ' (x) = 3 sin ^2 x cos x. We can write this function as:

f ' (x) = 3 \cdot sin ^2 x cos x

Now, we can easily apply the derivative product rule to find the second derivative of the function.  The derivative product rule says that m (x) \cdot n (x) = m ' (x) \cdot n(x) + n' (x) \cdot m (x).

The derivative of sin ^2 (x) is 2 sin ^x cos x and the derivative of cos x is - sin x. Substitute these derivative in the derivative product formula:

f '' (x) = 3 (  2 sin x cos x \cdot cos x + sin ^ 2 x \cdot - sin x)

f ''' (x) = 3 ( 2 sin ^2x cos x - sin ^3 x)

After simplification, we will get the following second derivative of the function:

f ''' (x) = 6 sin ^ 2 (x ) - 3 sin ^ 3 (x)

Example 5

Find the 1st, 2nd and 3rd derivatives of the function f (x) = (7x + 5) ^ 5.

Step 1 - First Derivative

We will use the derivative chain rule here to find the first derivative of the function. In the above function f(x) = x ^ 5 and g(x)  = 7x + 5. Hence, f(g (x) ) = (7x + 5 ) ^ 5.

f ' (x)= 5 \cdot (7x + 5) ^n{5 - 1} \cdot 7

f ' (x) = = 35 (7x + 5) ^ 4

Step 2 - Second Derivative

Again use the chain rule to find the next derivative. We will assume that f(x) = x ^ 4 and g (x) = 7x + 5.

f '' (x) = 35 \cdot 4 (7x + 5 ) ^ {4 - 1} \cdot 7

f '' (x) = 980 ( 7x + 5 ) ^ 3

Step 3 - Third Derivative

Use the chain rule to find the next higher order derivative. Again the chain rule will be used here. Now, we will assume that f(x) = x ^ 3 and g (x) = 7x + 5.

f ''' (x) = 980 \cdot 3 (7x + 5) ^ {3 - 1} \cdot 7

= 20580 ( 7x + 5 ) ^ 2

 

 

 

nth Derivative

Taking the derivatives of the function n number of times is known as nth derivative of the function. A general formula for all of the successive derivatives exists. This formula is called the nth derivative, f'n(x). It can be denoted as:

\frac {d ^ n} {d x ^ n}

Let us see the following example.

Calculate the nth derivative of the function f(x) = \frac {1} {x}.

f ' (x) = - \frac {1} {x ^ 2}

f '' (x) = \frac {1 \cdot 2} { x ^ 3} = \frac {2!} { x ^ 3}

f ''' (x) = - \frac {1 \cdot 2 \cdot 3} { x ^ 4} =- \frac {3!} { x ^ 4}

Hence, the formula for nth derivative of the function f(x) = \frac {1} {x} will be:

f ^ n (x) = (- 1) ^ n \frac {n !} { x ^ {n + 1}}

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.

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