Calculating Higher Order Derivatives

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  1. 0:06 Understanding the 'Jerk'
  2. 2:34 Finding the Third Derivative
  3. 5:20 Finding the Fourth Derivative
  4. 8:14 Lesson Summary
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Taught by

Erin Lennon

Erin has taught math and science from grade school up to the post-graduate level. She holds a Ph.D. in Chemical Engineering.

Differentiating functions doesn't have to stop with the first or even second derivative. Learn what a mathematical jerk is as you calculate derivatives of any order in this lesson.

Understanding the 'Jerk'

Derivatives and the rate of change
Derivatives and Rate of Change

I love riding on roller coasters. One of my favorite roller coasters of all time is Space Mountain. On this roller coaster you're kept in the dark the entire time. So picture it: you're going along in the dark, and suddenly you're jerked to one side, then jerked to the other as you careen around this Space Mountain. It's fantastic! But as I'm being jerked around, I always think about one thing: Why do we use the term 'jerk'? What in the world does that mean?

Well, it goes back to derivatives and the rate of change. If you have some function like y=f(t), the position, y, is a function of time, t. Say this is my height on the roller coaster. Then I can look at y`(dy/dt), which is the rate of change, d/dt, of my position, y. So this rate of change is my velocity, it's how fast my height is changing as a function of time. I could take the derivative of that, y``, or ((d^2)y)/dt^2, as the derivative of the rate of change of position, so it's the derivative of the velocity. And the derivative of the velocity is the acceleration. Well, the acceleration can also change, so I can write y```, and that's the rate of change of the acceleration. And how fast my acceleration changes is known as the jerk.

So you know how on a roller coaster you're completely stopped at first? There's no acceleration, no velocity, no nothing. Then all of a sudden you jerk, or lurch, forward. That's a change in your acceleration; that's d/dt of your acceleration. So the key here is that the derivative is just a rate of change. But in the real world, nothing is static. Everything is dynamic; everything changes. Static means stationary and unchanging, and dynamic means changing. You measure this change using derivatives.

Finding the jerk in the first example problem
Third Derivative Jerk

Finding the Third Derivative

Let's do an example. Let's say we have position, f(t), as a function of time, t, and it equals sin(t) + t^3. I know that the velocity is the derivative of the position, so f`(t) is d/dt sin(t) + t^3. That's my position. I can find this derivative by first dividing and conquering, so d/dt sin(t) + d/dt(t^3). Well using my derivative rules here, d/dt sin(t)=cos(t), and d/dt(t^3) is 3t^2, so my velocity is cos(t) + 3t^2. My acceleration is the derivative of the velocity - it's how fast my velocity is changing - and that's f``, or d/dt f`(t). I can calculate this by finding the derivative, d/dt, of my velocity, which is cos(t) + 3t^2. Again I can divide and conquer to get d/dt cos(t) + 3(d/dt)t^2. Then using my derivative rules, I find that the acceleration is -sin(t) + 3(2t), or -sin(t) + 6t. Now that I know the acceleration, I can find the jerk, which is just the derivative or the rate of change of the acceleration. This is f```, or d/dt f``(t), so that's d/dt of the acceleration, the rate of change of the acceleration. I can calculate that by finding d/dt of the acceleration, which is -sin(t) + 6t. Divide and conquer; that equals -(d/dt)sin(t) + 6(d/dt)t. Again using my rules I know that this equals -cos(t) + 6. So in this case, where my position was originally sin(t) + t^3, the jerk as a function of time - that's how fast my acceleration is changing as a function of time - is equal to -cos(t) + 6.

Use the chain rule to find the first derivative in the second example
Fourth Derivative Chain Rule

Finding the Fourth Derivative

We can use these same principles to find any higher-order derivative. So, for example, we can find the fourth-order derivative of f(x) = x^(-1) + cos(4x). This fourth-order derivative is f````. Mathematicians kind of get lazy after the first three, so we write f^4. Let's find the fourth-order derivative of this function, f(x) = x^(-1) + cos(4x).

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