Elianna Lawrence

Answered

2022-07-17

First Derivative vs Second Derivative in relation to minima and maxima

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?

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?

Answer & Explanation

taguetzbo

Expert

2022-07-18Added 16 answers

Step 1

You have a function f. A root, ${x}_{0}$ of the first derivative tells you where the extreme point is. ${f}^{\prime}({x}_{0})=0$.

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 ${f}^{\u2033}({x}_{0})$ will be negative.

Step 2

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.

You have a function f. A root, ${x}_{0}$ of the first derivative tells you where the extreme point is. ${f}^{\prime}({x}_{0})=0$.

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 ${f}^{\u2033}({x}_{0})$ will be negative.

Step 2

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.

Deborah Wyatt

Expert

2022-07-19Added 4 answers

Step 1

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 ${x}^{3}-3{x}^{2}+1$. If we are to use the "+" and "-" method, we must first consider the points on a line that we need to include. Since the $\frac{dy}{dx}=3{x}^{2}-6x=3x(x-2)=0$ 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 $-\mathrm{\infty}$ to 0, 0 to 2, and 2 to $\mathrm{\infty}$ are positive or negative by inputting values.

Step 2

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

Step 3

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.

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 ${x}^{3}-3{x}^{2}+1$. If we are to use the "+" and "-" method, we must first consider the points on a line that we need to include. Since the $\frac{dy}{dx}=3{x}^{2}-6x=3x(x-2)=0$ 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 $-\mathrm{\infty}$ to 0, 0 to 2, and 2 to $\mathrm{\infty}$ are positive or negative by inputting values.

Step 2

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

Step 3

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.

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