# Higher-order derivatives

We already know that how we can take a derivative of function, and we’ve learned how to find the first and second derivatives — but can we find a third derivative? Or even higher?

You bet we can! The process of differentiation can be performed several times in a row, which leads to higher-order derivatives.

## What does it mean to find a higher-order derivative?

A derivative of a function is a rate of change of a function with respect to a change of a variable. To find the higher-order derivative — or the $$n$$th derivative — means to find the derivative of the $$(n-1)$$th derivative of the function. So basically, it’s a derivative of a derivative of a derivative, etc.!

To help us during our differentiation, we’ll use differentiation rules rather than the definition of the derivative.

Constant multiple property of derivatives | $$\frac{d}{dx}\left(c\times f(x)\right)=c\times\frac{d}{dx}\left(f(x) \right)$$ |

Sum rule for derivatives | $$\frac{d}{dx}\left(f(x) + g(x)\right)=\frac{d}{dx}\left( f(x) \right)+\frac{d}{dx}\left( g(x) \right)$$ |

Difference rule for derivatives | $$\frac{d}{dx}\left(f(x) - g(x)\right)=\frac{d}{dx}\left( f(x) \right)-\frac{d}{dx}\left( g(x) \right)$$ |

Product rule for derivatives | $$\frac{d}{dx}\left(f(x)\times g(x)\right)=\frac{d}{dx}\left( f(x) \right)\times g(x)+f(x)\times\frac{d}{dx}\left( g(x) \right) $$ |

Quotient rule for derivatives | $$\frac{d}{dx}\left(\frac{f(x)}{g(x)} \right)=\frac{\frac{d}{dx}\left(f(x) \right)\times g(x)-f(x)\times\frac{d}{dx}\left( g(x) \right)}{(g(x))^{2}}$$ |

The Chain rule | $$(f\circ g)^{\prime}(x)=f^{\prime}(g(x))\times g^{\prime}(x) \text{ or } \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx} \text{ when }y=f(u), u=g(x)$$ |

Derivative of the inverse function | $$\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}\left(x\right)\right)}$$ |

### Why is the higher-order derivative so useful?

As we already know, the second derivatives can be really useful for graphing functions and determining the behavior of the preceding function’s graph!

## How to find the higher-order derivative

If you already have solid skills for taking the first and second derivatives, finding a higher-order derivative should just be a few extra steps for you! Either way, let’s walk through some detailed examples together.

### Example 1

**Find the higher-order derivative:**

To find the higher-order derivative (in this case the third derivative), we’ll differentiate three times

$$\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{d}{dx}\left(x^{7} + \frac{5}{3}x^{3} – 7x + \pi^{2}\right)\right)\right)$$

First, let’s find the first derivative:

$$\frac{d}{dx}\left(x^{7} + \frac{5}{3}x^{3} – 7x + \pi^{2}\right)$$

Use the differentiation rule $$\frac{d}{dx}(f+g)=\frac{d}{dx}(f)+\frac{d}{dx}(g)$$:

$$\frac{d}{dx}(x^{7}) + \frac{d}{dx}\left(\frac{5}{3}x^{3}\right) – \frac{d}{dx}(7x) + \frac{d}{dx}(\pi^{2}))$$

Use the differentiation rule $$\frac{d}{dx} (a\times f)=a\frac{d}{dx}(f)$$:

$$\frac{d}{dx}(x^{7}) + \frac{5}{3}\times\frac{d}{dx}(x^{3}) – 7\times \frac{d}{dx}(x) + \frac{d}{dx}(\pi^{2}))$$

Use the differentiation rule $$\frac{d}{dx} x^n=nx^{n-1}$$:

$$7\times x^{6} + \frac{5}{3}\times 3\times x^2 – 7\frac{d}{dx}(x) + \frac{d}{dx}(\pi^{2}))$$

Remember: the derivative of a variable to the first power is always $$1$$:

$$7\times x^{6} + \frac{5}{3}\times 3\times x^2 – 7\times 1 + \frac{d}{dx}(\pi^{2}))$$

The derivative of a constant is always $$0$$, so:

$$7\times x^{6} + \frac{5}{3}\times 3\times x^2 – 7\times 1 + 0$$

Removing the zero doesn’t change the value, so remove the zero:

$$7\times x^{6} + \frac{5}{3}\times 3\times x^2 – 7\times 1$$

Simplify the expression:

$$7 x^{6} + 5 x^2 – 7$$

On to the second derivative! Let’s take the derivative of each term, with respect to $$x$$:

$$\frac{d}{dx}(7 x^{6} + 5 x^2 – 7)$$

Use the differentiation rule $$\frac{d}{dx}(f+g)=\frac{d}{dx}(f)+\frac{d}{dx}(g)$$:

$$\frac{d}{dx}(7 x^{6} + 5 x^2 – 7)$$

Use the differentiation rule $$\frac{d}{dx}(f+g)=\frac{d}{dx}(f)+\frac{d}{dx}(g)$$:

$$\frac{d}{dx}(7 x^{6}) + \frac{d}{dx}(5 x^2) – \frac{d}{dx}(7)$$

Use the differentiation rule $$\frac{d}{dx} (a\times f)=a\frac{d}{dx}(f)$$:

$$7\times\frac{d}{dx}(x^{6}) + 5\times\frac{d}{dx}( x^2) – \frac{d}{dx}(7)$$

Use the differentiation rule $$\frac{d}{dx} x^n=nx^{n-1}$$:

$$7\times6x^5 + 5\times2x – \frac{d}{dx}(7)$$

The derivative of a constant is always $$0$$:

$$7\times6x^5 + 5\times2x – 0$$

Removing the zero doesn’t change the value, so we’ll get rid of that zero:

$$7\times6x^5 + 5\times2x$$

Simplify the expression:

$$42x^5 + 10x$$

Here we go: time for the third derivative! Let’s differentiate the expression one more time:

$$\frac{d}{dx}(42x^5 + 10x)$$

Use the differentiation rule $$\frac{d}{dx}(f+g)=\frac{d}{dx}(f)+\frac{d}{dx}(g)$$:

$$\frac{d}{dx}(42x^5) + \frac{d}{dx}(10x)$$

Use the differentiation rule $$\frac{d}{dx} (a\times f)=a\frac{d}{dx}(f)$$:

$$42\times\frac{d}{dx}(x^5) + 10\times\frac{d}{dx}(x)$$

Use the differentiation rule $$\frac{d}{dx} x^n=nx^{n-1}$$:

$$42\times5\times x^4 + 10\times\frac{d}{dx}(x)$$

The derivative of a variable to the first power is always $$1$$:

$$42\times5\times x^4 + 10\times1$$

Simplify the expression:

$$210x^4 + 10$$

Woohoo! We found the third derivative of $$x^{7} + \frac{5}{3}x^{3} – 7x + \pi^{2}$$:

Once you get into the swing of things, it’s not so bad! Let’s look at another example.

### Example 2

**Find the higher-order derivative:**

To find the higher-order derivative (in this case the third derivative), take the derivative three times:

$$\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{d}{dx}\left(x^2+3\right)\right)\right)$$

First up is the first derivative:

$$\frac{d}{dx}\left(x^2+3\right)$$

Use the differentiation rule $$\frac{d}{dx}(f+g)=\frac{d}{dx}(f)+\frac{d}{dx}(g)$$:

$$\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(3\right)$$

Use the differentiation rule $$\frac{d}{dx} x^n=nx^{n-1}$$:

$$2x+\frac{d}{dx}\left(3\right)$$

The derivative of a constant is always $$0$$:

$$2x+0$$

Removing the zero doesn’t change the value, so remove the zero:

$$2x$$

Great! Now to find the second derivative, take the derivative of each term, with respect to $$x$$:

$$\frac{d}{dx}(2x)$$

Use the differentiation rule $$\frac{d}{dx} (a\times f)=a\frac{d}{dx}(f)$$:

$$2\times\frac{d}{dx}(x)$$

The derivative of a variable to the first power is always $$1$$:

$$2\times1$$

Multiply the numbers:

$$2$$

So close! Take the derivative one more time so we can find our third derivative:

$$\frac{d}{dx}(2)$$

The derivative of a constant is always $$0$$:

$$0$$

Got it! The third derivative of $$x^2+3$$ is:

Finding higher-order derivatives can be repetitive, but sometimes that makes things easy! Remember these steps when you try this process on your own:

## Study summary

- To find the higher-order derivative, take the derivative multiple times.
- Find the derivative.
- To find the n-th derivative, repeat the process n times.
- Simplify the final expression, if possible.

## Do it yourself!

Looking for some more examples? Try these practice problems and see how you do!

**Take the derivative of a function:**

- $$\frac{d^4}{dx^4}\left(e^x + 5x\right)$$
- $$\frac{d^2}{dx^2}\left(\ln{(x)}+5^x\right)$$
- $$\frac{d^3}{dt^3}\left(\sqrt{t^4-1}\right)$$
- $$\frac{d^3}{dx^3}\left(\frac{x-1}{x^2-4}\right)$$

*Solutions:*

- $$e^x$$
- $$-\frac{1}{x^2}+\ln(5)^2\times5^x$$
- $$\frac{12t^5+12t}{\sqrt{t^4-1}(t^4-1)^2}$$
- $$\frac{-6x^4+24x^3-144x^2+96x-96}{(x^2-4)^4}$$

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