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I am struggling with understanding the difference between them and I need to write about them intuitively. The way my teacher explained it, the sign of the first derivative is used to determine if there is a minimum, maximum, or neither. For example, if the derivative is increasing to the left of x and decreasing to the right of x, than a maximum is present. If the derivative is decreasing to the left of x and is increasing to the right of x, there is a minimum. I know that the second derivative is the derivative if the first derivative and if it is positive on a certain interval, it is concave up. Is that not telling me the same thing as the first derivative?

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  • $\begingroup$ Just think about what the derivative means, the first derivative of a function is its slope. The second derivative is the slope of the original functions derivative, or the concavity of $f$. As long as you understand the relationships of these functions to each other you should be able to solve problems that move between them. $\endgroup$
    – EgoKilla
    Dec 10, 2014 at 22:46
  • $\begingroup$ It does not tell the same thing because a function can be increasing on an interval and also change convexity. Consider $f(x) = \arctan x$. $\endgroup$
    – rubik
    Dec 10, 2014 at 22:46

5 Answers 5

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The second derivative tells you about the first derivative what the first derivative tells you about the function you're deriving. You can tell plenty of things about a function from the first derivative, but it's always better to take the second derivative to study the function in a much proper and exact way.

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  • $\begingroup$ So when I'm writing, would it be accurate to say that if the value of the second derivative is positive, than it is concave up, so a minimum exists and if it is negative, than it is concave down, so a maximum exists? $\endgroup$
    – Sara
    Dec 10, 2014 at 22:46
  • $\begingroup$ Yes, that's exactly what the second derivative is for (one of its applications). Actually if there is a minimum at $x_0$ its way better to write $x_0$ is a minimum because it is a critical point and $f''(x_0)>0$, than writing: well, we see that the first derivative increases from... It's just a more formal way of telling it. $\endgroup$ Dec 10, 2014 at 22:46
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Suppose that $f(x)$ has a minimum at $x_0$. Then, of course, you can think of the point $x_0$ as intuitively being the bottom of some sort of dip in the function's graph.

The first derivative $f'(x)$ will always be zero at $x_0$ - that is, $f'(x_0)=0$. Why is this? It's simply because the derivative gives the slope of a tangent, and the slope of a tangent to the bottom of a cup or dip is zero - the tangent itself is a horizontal line. This also works at a maximum - however like you said, if the derivative is decreasing to the left and increasing to the right of $x_0$ then it's a minimum and vice versa means it's a maximum. That part is all right.

Now, consider the graph of the derivative. All we know is that the graph crosses the $x$ axis at $x=0$ from our discussion before. The second derivative $f''(x)$ will be positive at this point if the function is at a minimum and negative if the function is at a maximum. Hence you're getting the same information - it's just easier to tell whether a function is positive or negative than whether it's increasing or decreasing. Differentiating gives a nice way of doing just that!

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What you say about the first derivative is true, and, yes, the second derivative gives information about concavity as well. Why is it important to have the second derivative test?

When we have a complicated function, it is usually easier to evaluate the second derivative to determine maxima and minima than it is creating a line and plotting "-" and "+" everywhere. If the second derivative is equal to zero, we must use the foundational method of plotting the "-" and "+" between specified points to determine if we have a maxim, minima, or constant between points.

As an example, say we have $x^3-3x^2+1$. If we are to use the "+" and "-" method, we must first consider the points on a line that we need to include. Since the $\frac{dy}{dx} = 3x^2-6x =3x(x-2) = 0$ we have two points for $3x=0$ and $(x-2) = 0$. Solving the latter and we get $x = 0$ and $x=2$. We now must create a line and determine if the intervals between $-\infty$ to $0$, $0$ to $2$, and $2$ to $\infty$ are positive or negative by inputting values.

In contrast, with the second derivative test, all we have to do is take the second derivative: $\frac{d^2y}{dx^2}$ = 6x-6. The curve is concave up when the second derivative is greater than $0$ and concave down when the derivative is less than $0$.

Concave up:

$$6x-6 > 0$$ $$x > 1$$

Concave down:

$$6x-6 < 0$$ $$x < 1$$

There is much less work required. If the second derivative is zero, one must use the first derivative method.

References:

[1] Anton, Howard. (1992). Calculus with analytic geometry-4th Edition. Anton Textbooks, Inc.

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In this case, yes it is telling you the same thing. Since you knew you had a point where the first derivative was zero and it was positive to one side and negative to the other, you already knew the concavity.

But sometimes you merely find the zeroes of the first derivative and don't know that the sign of the first derivative is to the left or right. If you just have the zeroes, you can check the 2nd derivative to determine the concavity.

This is specific to the case you were asking about. There is other information you can determine about your original function using the two derivative together.

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You have a function $f$. A root, $x_0$ of the first derivative tells you where the extreme point is. $$f'(x_0) = 0 $$ But not if it is maximum or minimum. So, indeed, then you can check what is going on around $x_0$ in $f$. If it is increasing at the left and decreasing at the right, then you have a maximum. But if this happens, then the first derivative $f'$, will be positive at the left and negative at the right of $x_0$. But then, this means that the slope of the first derivative is negative (because it is decreasing), so the sign of $f''(x_0)$ will be negative.

Conclusion, wether you check the slope of the first derivative at a root, or the sign of the second derivative at the root, it is the same thing.

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