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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