Using d^ny/dx^n to Denote the nth Derivative of y=f(x)

How to use {eq}\frac{\mathrm{d^{n}} y}{\mathrm{d} x^{n}} {/eq}to Denote the nth Derivative of y=f(x): Step by Step Overview

Step 1: Understand that the derivative of any function with respect to x is referring to the slope of that function at x.

Recall that the slope formula is $$m=\frac{\Delta y}{\Delta x} $$

Now {eq}\Delta {/eq} as a symbol is generally used to denote "large" changes in a variable. When talking about derivatives though, by definition we are talking about infinitesimally small changes in a variable. This is when we switch to a slightly different notation. Instead of {eq}\Delta {/eq} we simply use a lower-case d. So we make the switch from {eq}\frac{\Delta y}{\Delta x} {/eq} to {eq}\frac{dy}{dx} {/eq}.

The symbol we use to indicate that we are performing the operation of "taking the derivative" is {eq}\frac{\mathrm{d} }{\mathrm{d} x} {/eq}.

So if we have {eq}y=f(x) {/eq}, taking the derivative will look like this:$$\frac{\mathrm{d} }{\mathrm{d} x}y=\frac{\mathrm{d} }{\mathrm{d} x}f(x) $$

Also, the derivatives of a function are often denoted as$$f^{'}(x),f^{''}(x),f^{'''}(x),f^{(4)}(x),f^{(5)}(x),..... $$

So upon taking the derivative of our function, we would write:$$\frac{\mathrm{d} y}{\mathrm{d} x}=f^{'}(x) $$

Step 2: Using the notation from Step 1, continuously take the derivative of your function until you reach the nth derivative.

First Derivative: $$\frac{\mathrm{d} }{\mathrm{d} x}y=\frac{\mathrm{d} }{\mathrm{d} x}f(x)\rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=f^{'}(x) $$

Second Derivative: $$\frac{\mathrm{d} }{\mathrm{d} x}(\frac{\mathrm{d} y}{\mathrm{d} x}) = \frac{\mathrm{d} }{\mathrm{d} x}[f^{'}(x)]\rightarrow \frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=f^{''}(x) $$

Third Derivative: $$\frac{\mathrm{d} }{\mathrm{d} x}(\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}) = \frac{\mathrm{d} }{\mathrm{d} x}[f^{''}(x)]\rightarrow \frac{\mathrm{d^{3}} y}{\mathrm{d} x^{3}}=f^{'''}(x) $$

nth Derivative: $$\frac{\mathrm{d} }{\mathrm{d} x}(\frac{\mathrm{d^{n-1}} y}{\mathrm{d} x^{n-1}}) = \frac{\mathrm{d} }{\mathrm{d} x}[f^{(n-1)}(x)]\rightarrow \frac{\mathrm{d^{n}} y}{\mathrm{d} x^{n}}=f^{(n)}(x) $$

It is also worth noting a couple of things about this choice of notation:

• It is especially useful when working with Differential Equations as it is easier to manipulate.
• It also tells/reminds us that the derivative is simply a rate of two numbers. More specifically, it is the slope at any particular point.

Using {eq}\frac{\mathrm{d^{n}} y}{\mathrm{d} x^{n}} {/eq} to Denote the nth Derivative of y=f(x): Relevant Rules and Properties of Derivatives

Below are just some of the basic rules for derivatives. We will be using these in the following examples.

Constant Rule: {eq}\frac{\mathrm{d} }{\mathrm{d} x}c = 0 {/eq}

Constant Multiple Rule: {eq}\frac{\mathrm{d} }{\mathrm{d} x}cf(x) = c\frac{\mathrm{d} }{\mathrm{d} x}f(x) {/eq}

Power Rule: {eq}\frac{\mathrm{d} }{\mathrm{d} x}x^{n} = nx^{n-1} {/eq}

Sum and Difference Rule: {eq}\frac{\mathrm{d} }{\mathrm{d} x}[f(x)\pm g(x)] = \frac{\mathrm{d} }{\mathrm{d} x}f(x)\pm \frac{\mathrm{d} }{\mathrm{d} x}g(x) {/eq}

Product Rule: {eq}\frac{\mathrm{d} }{\mathrm{d} x}[f(x)g(x)] = f(x)[\frac{\mathrm{d} }{\mathrm{d} x}g(x)]+g(x)[\frac{\mathrm{d} }{\mathrm{d} x}f(x)] {/eq}

Quotient Rule: {eq}\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)} = \frac{g(x)[\frac{\mathrm{d} }{\mathrm{d} x}f(x)]-f(x)[\frac{\mathrm{d} }{\mathrm{d} x}g(x)]}{[g(x)]^{2}} {/eq}

Chain Rule: {eq}\frac{\mathrm{d} }{\mathrm{d} x}f(g(x)) = [\frac{\mathrm{d} }{\mathrm{d} x}f(g(x))]\frac{\mathrm{d} }{\mathrm{d} x}g(x) {/eq}

Also, we will be using the following well-known derivatives of Sine and Cosine:

• {eq}\frac{\mathrm{d} }{\mathrm{d} x}sin(x) = cos(x) {/eq}

• {eq}\frac{\mathrm{d} }{\mathrm{d} x}cos(x) = -sin(x) {/eq}

The following examples demonstrate how to use this notation when given a particular function.

Example 1: Finding the 3rd Derivative of a Polynomial

In this example, let's look at the following polynomial:

$$y=4x^{6}+x^{4}-2x^{2} $$

We're trying to find the 3rd derivative, so we are looking to find {eq}\frac{\mathrm{d^{3}} y}{\mathrm{d} x^{3}} {/eq}.

Taking the 1st derivative:

$$\frac{\mathrm{d} }{\mathrm{d} x}y=\frac{\mathrm{d} }{\mathrm{d} x}(4x^{6}+x^{4}-2x^{2}) $$

Using the Sum/Difference Rule, we get:$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}4x^{6}+\frac{\mathrm{d} }{\mathrm{d} x}x^{4}-\frac{\mathrm{d} }{\mathrm{d} x}2x^{2} $$

Then using the Constant Multiple Rule, we can rewrite the function:$$\frac{\mathrm{d} y}{\mathrm{d} x}=4\frac{\mathrm{d} }{\mathrm{d} x}x^{6}+\frac{\mathrm{d} }{\mathrm{d} x}x^{4}-2\frac{\mathrm{d} }{\mathrm{d} x}x^{2} $$

Now using the Power Rule to evaluate the derivative, we get:

$$\frac{\mathrm{d} y}{\mathrm{d} x}=4(6)x^{6-1}+(4)x^{4-1}-2(2)x^{2-1} $$

Simplifying gives us:

$$\frac{\mathrm{d} y}{\mathrm{d} x}=24x^{5}+4x^{3}-4x $$

At this point, we have found the 1st derivative. Now we take the derivative two more times to get to the 3rd derivative.

$$\frac{\mathrm{d} }{\mathrm{d} x}(\frac{\mathrm{d} y}{\mathrm{d} x})=\frac{\mathrm{d} }{\mathrm{d} x}(24x^{5}+4x^{3}-4x) $$

$$\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=\frac{\mathrm{d} }{\mathrm{d} x}24x^{5}+\frac{\mathrm{d} }{\mathrm{d} x}4x^{3}-\frac{\mathrm{d} }{\mathrm{d} x}4x $$

$$\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=24\frac{\mathrm{d} }{\mathrm{d} x}x^{5}+4\frac{\mathrm{d} }{\mathrm{d} x}x^{3}-4\frac{\mathrm{d} }{\mathrm{d} x}x $$

$$\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=24(5)x^{5-1}+4(3)x^{3-1}-4(1)x^{1-1} $$

$$\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=96x^{4}+12x^{2}-4 $$

Finally, we find the 3rd.

$$\frac{\mathrm{d} }{\mathrm{d} x}(\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}})=\frac{\mathrm{d} }{\mathrm{d} x}(96x^{4}+12x^{2}-4) $$

$$\frac{\mathrm{d^{3}} y}{\mathrm{d} x^{3}}=\frac{\mathrm{d} }{\mathrm{d} x}96x^{4}+\frac{\mathrm{d} }{\mathrm{d} x}12x^{2}-\frac{\mathrm{d} }{\mathrm{d} x}4 $$

In this next step, use the Constant Multiple Rule for the last term.

$$\frac{\mathrm{d^{3}} y}{\mathrm{d} x^{3}}=96(4)x^{4-1}+12(2)x^{2-1}-0 $$

After simplifying, we have our 3rd derivative:

$$\frac{\mathrm{d^{3}} y}{\mathrm{d} x^{3}}=384x^{3}+24x $$

Example 2: Finding the 4th Derivative of a Trigonometric Function

Let's take a look at the following function:

$$y=sin(2x) $$

We're trying to find the 4th derivative, so we are looking to find {eq}\frac{\mathrm{d^{4}} y}{\mathrm{d} x^{4}} {/eq}.

Taking the 1st derivative:

$$\frac{\mathrm{d} }{\mathrm{d} x}y=\frac{\mathrm{d} }{\mathrm{d} x}sin(2x) $$

Using the Chain Rule, we get:

$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}sin(2x)\frac{\mathrm{d} }{\mathrm{d} x}(2x) $$

Using the derivative of Sine:

$$\frac{\mathrm{d} y}{\mathrm{d} x}=2cos(2x) $$

This is the first derivative. Let's continue to take three more derivatives to find the 4th one.

$$\frac{\mathrm{d} }{\mathrm{d} x}(\frac{\mathrm{d} y}{\mathrm{d} x})=\frac{\mathrm{d} }{\mathrm{d} x}2cos(2x) $$

$$\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=2\frac{\mathrm{d} }{\mathrm{d} x}cos(2x) $$

Using the derivative of Cosine and the Chain Rule:

$$\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=-2sin(2x)\frac{\mathrm{d} }{\mathrm{d} x}(2x) $$

$$\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=-4sin(2x) $$

Continue to take two more derivatives:

$$\frac{\mathrm{d} }{\mathrm{d} x}(\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}})=\frac{\mathrm{d} }{\mathrm{d} x}(-4sin(2x)) $$

$$\frac{\mathrm{d^{3}} y}{\mathrm{d} x^{3}}=-4\frac{\mathrm{d} }{\mathrm{d} x}sin(2x)\frac{\mathrm{d} }{\mathrm{d} x}(2x) $$

$$\frac{\mathrm{d^{3}} y}{\mathrm{d} x^{3}}=-8cos(2x) $$

Now for the last derivative:

$$\frac{\mathrm{d} }{\mathrm{d} x}(\frac{\mathrm{d^{3}} y}{\mathrm{d} x^{3}})=\frac{\mathrm{d} }{\mathrm{d} x}(-8cos(2x)) $$

$$\frac{\mathrm{d^{4}} y}{\mathrm{d} x^{4}}=-8\frac{\mathrm{d} }{\mathrm{d} x}cos(2x)\frac{\mathrm{d} }{\mathrm{d} x}(2x) $$

$$\frac{\mathrm{d^{4}} y}{\mathrm{d} x^{4}}=16sin(2x) $$

So the fourth Derivative of {eq}y=sin(2x) {/eq} is {eq}\frac{\mathrm{d^{4}} y}{\mathrm{d} x^{4}}=16sin(2x) {/eq}.