Higher Order Derivatives - Online Tutor, Practice Problems & Exam Prep

Higher Order Derivatives

If I asked you to find the derivative of this polynomial function, you could do so using all of the rules that we've learned. But what if I asked you to find the second derivative of this function? What does that even mean? Well, the second derivative is what we would refer to as a higher order derivative. And all higher order derivatives are just repeated differentiation, meaning all we have to do is just take the derivative again using all of the rules that we already know.

So here I'm going to show you exactly how to deal with higher order derivatives, whether it's the second derivative, the 4th derivative, or even the 100th derivative. So let's go ahead and jump right in here. Now, I just said that higher order derivatives are just repeated differentiation. So, to find the second derivative of a function, we just need to take the derivative twice. Now here, we've already taken the derivative once.

That is our first derivative that we've already seen before, so we just need to take the derivative again, differentiating this d/dx (3x2-2x+5). Now taking the derivative of that first term using the power rule, the derivative of 3x2 is just 6x. Then the derivative of a negative 2x is a negative 2. And then the derivative of that constant 5 is just 0, so I don't have to worry about it.

And this is my second derivative, f′′⁡(x), using 2 little prime marks to denote that this is the second derivative. Now we can take this even further and find our 3rd derivative by just taking the derivative again. Now taking the derivative of this 6x-2 will just give me 6 for that 3rd derivative, which I can denote with 3 little prime marks, f′′′⁡(x), for that third derivative.

Now we can take this even further and go ahead and find our 4th derivative just differentiating that 6. Since that is a constant, I'm just going to get 0 for that 4th derivative. Now instead of just continually adding more prime marks, at this point, we can denote this just with a little 4 in parentheses up here to show that this is the 4th derivative. Now we could take this even further because we saw here that to find the second derivative, we took the derivative twice. And to find the 4th derivative, we took the derivative 4 times. So to find the 100th derivative, we could take the derivative 100 times. But luckily, you shouldn't be asked to do that.

Now you may encounter a couple of different notations when dealing with these higher order derivatives. Now we already saw this first notation here with our little n in parentheses just denoting what order derivative we're on. Like, here we saw with our 4th derivative. Now you may also see this capital D again with this n just telling you what derivative you're on as well as these other two notations with n still denoting what order derivative you're on. So now that we know how to deal with these higher order derivatives, let's get some practice with them.

I'll meet you in the next video.

Higher Order Derivatives Example 1

In this problem, we're asked to find \( f''(x) \) or the second derivative. Now, our function that we're given here is \( f(x) = 4x^{-3} + 3x^7 \). Remember that in order to find the second derivative, it just means that we need to take the derivative twice. So let's start by just taking the derivative once, and then we can find it the second time to get our final answer. Now to find our first derivative here, \( f'(x) \), I am, of course, going to use the power rule since I have these powers happening here.

So, for that first term, \( 4x^{-3} \), I'm going to pull that negative 3 out to the front, multiplying by it, and then decrease that power by 1. So, that negative 3 times 4 will give me a negative 12, and then I have \( x^{-3 - 1} \), which is \( x^{-4} \). Then for that second term, I have \( 3x^7 \). Again, using the power rule here, I'm going to pull that 7 out to the front, multiplying by it, and then decrease that exponent by 1. So, that 3 times 7 gives me a positive 21, and then I have \( x^{7 - 1} \), which is \( x^6 \).

Now, we're not done yet because this is just our first derivative, and we want to find our second derivative, \( f''(x) \), by taking the derivative one more time. Taking the derivative of this function here again using the power rule just for a second time. I'm going to pull that negative 4 out to the front, multiplying by it, and then decrease that power by 1. Now, here, since I have negative 12 times negative 4, that's going to end up giving me a positive 48. So I have \( 48x^{-4 - 1} \), which is \( 48x^{-5} \).

Then for that second term, using the power rule again, pulling that 6 out to the front, multiplying it by it, that will give me 126. And then decreasing the power of that \( x \) by 1, that will give me \( x^{6 - 1} \), which is \( x^5 \). And this is my final answer here. The second derivative, \( f''(x) \), is \( 48x^{-5} + 126x^5 \). Let me know if you have any questions here.

Thanks for watching.

Higher Order Derivatives Example 2

In this problem, we're told that for our function \( f(x) \), we want to find the 5th derivative of this function. Now the function that we're given here is \( f(x) = x^8 \). Now remember that to find the 5th derivative, that just means we need to take the derivative five times. So let's go ahead and jump in here. Now in order to get to our 5th derivative, we, of course, have to take our 1st, 2nd, 3rd, and 4th derivatives.

So let's start with that first derivative \( f'(x) \). Now \( f'(x) \), if I'm using the power rule here, I want to pull that 8 to the front and then decrease that power by 1. This will give me \( 8x^7 \). Now in order to take our second derivative here, \( f''(x) \), I'm just going to use the power rule yet again, pulling that 7 out to the front, then decreasing that power by 1. Now this will give me \( 56x^6 \).

We can see that our power is just going down by 1 every time, which makes sense because we're using the power rule. Now for our 3rd derivative \( f'''(x) \), again, pulling that power out to the front, 56 times 6, and then decreasing that power by 1. This will give me \( 336x^5 \). Now we're going to move on to our 4th derivative. This is where instead of using primes, we're actually just going to put that number in parentheses up there.

So I have \( f^{(4)}(x) \). And then again, using the power rule, pulling that 5 out to the front, multiplying by it, and decreasing that power by 1. Now that's 336 times 5. We see that this constant is getting really large. This is going to give me \( 1680x^4 \).

One final derivative to take to get our final answer, our 5th derivative. So in taking our 5th derivative here, again, putting that 5 in parentheses with my higher order derivative notation. Again, using the power rule one final time, pulling that 4 out to the front, multiplying by it, decreasing that power by 1, this will give me \( 6720x^3 \).

So here, my 5th derivative of my function \( f(x) \) is \( 6720x^3 \). Now feel free to let me know if you have any questions here, and let's continue getting practice with higher order derivatives.

Find the third derivative of the given function.

$f(x)=24+4x^5$

• A. $60x^2$
• B. $80x^3$
• C. $240x^2$
• D. $24-240x^2$

Find the third derivative of the given function.

$y=3x^2+9x+1$

• A. $0$
• B. $6$
• C. $x$
• D. $3x^2$

Find the third derivative of the given function.

$f(t)=5t^{-4}$

• A. $0$
• B. $120t$
• C. $600t^7$
• D. $600t^{-7}$