Finding first and second derivative

So I'll just give you a sample of what can be done.s(t) = (1/3)*x^3 - 2x^2 + 3x + 8s'(t) = x^2 - 4x + 3s''(t) = 2x - 4time | position | velocity | acceleration | speed0 | 8 | 3 | -4 | 31 | 9 1/3 | 0 | -2 | 02 | 8 2/3 | -4 | 0 | 43 | 8 | -4 | 2 | 4
  • #1
diye
4
0
how would I find the first and second derivative of

(x^3/3) - 2x^2 + 3x +8

thanks
 
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  • #2
By differentiating (x^3/3) - 2x^2 + 3x +8,
and then differentiating what you got from the previous step.

What have you tried?
 
  • #3
thanks for the response, I am just having trouble with the understanding the fraction (x^3/3) and how to find the derivative. I can do the rest fine if I can figure out how to get the derivative of (x^3/3).

could I do this, multiply the whole equation by 3 to get rid of the fraction? Can someone be kind enough to show me how to find the derivative of a fraction with an x variable int he numerator? I know something like 3/x^3 is rewritable as 3^-3 but that has the x in the denominator.

thanks
 
  • #4
You don't really have an equation, so you can't multiply it by 3.

x^3/3 is the same as 1/3 * x^3. Do you know that d/dx(a*f(x)) = a * d/dx(f(x))?

BTW, 3/(x^3) = 3*x^(-3)
 
  • #5
yes I understand how and have found the derivative of (x^3/3) is x^2 but don't understand how exactly one gets to that. Do you just take the derivative of x^3 and get x^2 and then derivative of 3 which is 0 so you get x^2/0 = 0? that's what I was trying to do but don't see how you get x^2, do you just ignore the constant?

thanks
 
  • #6
If you just follow the formula (d/dx)x^n=n*x^(n-1). So for example (1/3)*x^3.
(d/dx)(1/3)*x^3
=(1/3)*3*x^2 (the constant can be pulled out front)
= x^2
 
  • #7
thanks, this is actually a rectilinear motion problem

alright, now the next step, original problem is

f(x) = x^3/3-2x^2 + 3x +8

first derivative: x^2 - 4x + 3 ... x = 3 x = 1
second derivative: 2x - 4 ... x = 2

s(0) = 8
s(1) = 9 1/3
s(2) = 8 2/3
s(3) = 8

so now

time | velocity | acceleration | speed
0 < x <1
1< x < 2
2< x < 3
x> 3

how would I find out the velocity acceleration speed?
 
  • #8
diye said:
f(x) = x^3/3-2x^2 + 3x +8

first derivative: x^2 - 4x + 3 ... x = 3 x = 1
second derivative: 2x - 4 ... x = 2
You have too much stuff.
f'(x) = x^2 - 4x + 3 PERIOD
f''(x) = 2x - 4 PERIOD

Now if you wanted to solve the equation f'(x) = 0, which you said nothing about, you would get x = 3 or x = 1. But the first derivative, or f'(x) is just x^2 - 4x + 3.

Similarly for the second derivative: f''(x) = 2x -4.
diye said:
s(0) = 8
s(1) = 9 1/3
s(2) = 8 2/3
s(3) = 8

so now

time | velocity | acceleration | speed
0 < x <1
1< x < 2
2< x < 3
x> 3

how would I find out the velocity acceleration speed?

You didn't identify a function with a letter in your first post, but it now appears that what you started with was s(t) = 1/3 * x^3 - 2x^2 + 3x + 8.

The first and second derivatives, s'(t) and s''(t) are what you found.
If s(t) is the position at time t, s'(t) is the velocity, and s''(t) is the acceleration. The speed is |s'(t)|.

Apparently you have to fill out a table, but I don't know what's supposed to go into the table. You can't put an interval (e.g., 0 < x < 1) into a function.
 

1. What is the purpose of finding the first and second derivative?

The first and second derivative are used to calculate the rate of change and the curvature of a function, respectively. They are important tools in understanding the behavior of a function and can be used to solve various optimization problems.

2. How do you find the first derivative of a function?

To find the first derivative of a function, you can use the power rule, product rule, quotient rule, or chain rule. These rules involve taking the derivative of each term in the function and combining them using algebraic operations.

3. What does the first derivative tell us about a function?

The first derivative tells us about the slope of the function at a specific point. It can also be used to determine whether the function is increasing or decreasing at that point, as well as the location of any maximum or minimum points.

4. How do you find the second derivative of a function?

The second derivative can be found by taking the derivative of the first derivative. In other words, you can apply the same rules used to find the first derivative to the first derivative itself. The resulting second derivative can tell us about the concavity and inflection points of the function.

5. What is the significance of the second derivative in calculus?

The second derivative is important in calculus as it provides information about the rate of change of the slope (or the curvature) of a function. It can be used to determine the concavity of a function, which is crucial in optimization problems. Additionally, the second derivative test can be used to find the nature of a critical point, whether it is a minimum, maximum, or inflection point.

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