4.3: Second Derivatives and Interpreting the Derivative
We understand velocity versus position at this point, but what about the derivative of other changing quantities? What does it mean? Let’s look at an example.
Let \(M(t)\) be the mass in metric tonnes of a glacier over a given time \(t\), where \(t\) is measured in years. What do you think the graph of \(M(t)\) looks like? Given global warming, it’s probably going down, like this:
Question: What is the derivative measuring now? What would the graph of the derivative look like?
Answers: Good questions! Just like the derivative of position is velocity, or how fast the position is changing, the derivative of \(M(t)\) is going to be how fast the mass of the glacier is changing. In this case, it is how fast it is melting.
We can automatically assign units to the derivative. Since the original graph is metric tonnes on the \(y\)-axis, and years on the \(x\)-axis, we know the unit of the derivative (unless we want to convert) is going to be metric tonnes per year. We might then talk about the derivative being \(-15 \ \frac{\text{metric tonnes}}{\text{year}}\).
Notice I assumed the derivative was negative. Why did I do that? That is because the graph is going down, and the glacier is losing mass.
Note further that it isn’t just going down like a line going down. It’s a curve downward. What does this mean for the derivative? Well, the derivative starts negative since it’s going down, and continues to be negative as it goes down. But it goes down faster and faster as you move to the right — that means the derivative is getting more and more negative. Like this:
So we can see that this derivative is negative, but it’s worse than that — it’s negative and going down. That’s not good news for the glacier. Looking at whether the derivative is going up or down is known as the second derivative . We will see in the next section this is easy to calculate.
As we saw in the above example, sometimes we need to repeat the process of taking derivatives. This gives the second derivative, third derivative, and so on. The notation is
\(f''(x), \quad \text{or} \quad \frac{d^2}{dx^2}f\)
You can also take three, four, or more derivatives. Instead of writing several primes, we write \(f^{(4)}\) for the fourth derivative, \(f^{(5)}\) for the fifth derivative, and so on. Let’s do a couple of examples.
- \(f'(x)\)
- \(f''(x)\)
- \(f'''(x)\)
- \(f^{(4)}(x)\)
For (1), we use the power rule and see that \(f'(x) = \boxed{3x^2 + 2x}\).
For (2), we use the power rule again applied to \(3x^2 + 2x\). So we have
\(f''(x) = 3 \frac{d}{dx} x^2 + 2 \frac{d}{dx} x = 3(2x) + 2(1) = \boxed{6x + 2}\)
For (3), we take the derivative of \(6x + 2\). This is \(\boxed{6}\).
For (4), we take the derivative of \(6\) and that is \(\boxed{0}\).
To get a second derivative, we first need to take one derivative. We see the derivative of \(\ln(x)\) is \(\frac{1}{x}\). We then take the derivative again, and we see
\[\begin{align*} \frac{d^2}{dx^2} \ln(x) & = \frac{d}{dx} \frac{1}{x} \\ & = \frac{d}{dx} x^{-1} \\ & = -1 x^{-2} \\ & = \boxed{-\frac{1}{x^2}} \end{align*}\]
Here, we see that \(f'(x) = -\sin(x)\), \(f''(x) = -\cos(x)\), \(f'''(x) = \sin(x)\), and \(f^{(4)}(x) = \cos(x)\). Hence, we start repeating answers after we take four derivatives!