# How can we calculate this derivative?

y = x.(2x^3-1)^4

y'' = ?

#### Cervesa

product rule and the chain rule

$y' = x \cdot 4(2x^3-1)^3 \cdot 6x^2 + (2x^3-1)^4 \cdot 1$

$y' = (2x^3-1)^3 \cdot [24x^3 + (2x^3-1)] = (2x^3-1)^3 \cdot (26x^3-1)$

finish by using product and chain rules again on $y'$ to find $y''$

topsquark

#### Prove It

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Just out of interest, why is a derivatives question posted in Pre-Calculus?

#### Monoxdifly

Just out of interest, why is a derivatives question posted in Pre-Calculus?
"Pre-University Math Help"

#### Prove It

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"Pre-University Math Help"
A Calculus question in "Pre-Calculus"...

#### hbwebsol

Derivatives
Suppose you've just watched a car race on an out-and-back course. The drivers drove 2,800 feet out and 2,800 feet back. The winner of the race drove in such a way that her distance from the start can be modeled using the function:
f(x) = -7x2 + 280x
where x is the number of seconds since the start of the race.
As she was bragging about first place, someone asked her how fast she was going. She realized that she knew her speed was different at different points in the race, but she wasn't sure how to find how fast she was going at any given point. Hmmm... any ideas? Thankfully, there is a mathematical answer to this conundrum, and that answer lies in derivatives.
The derivative of a function is the rate at which the function value is changing, with respect to x, at a given value of x. Therefore, if we can find the derivative of the winner's distance function, then we can find how fast she was going at any given time in the race. Let's take a look at how to do this!

Computing Derivatives
You may recall something called the difference quotient from an algebra or pre-calculus course. The difference quotient of a function f(x) is a formula that gives the slope of the line through any two points with x-coordinates x and x + h on the function:
(f(x + h) - f(x)) / h
This is the key to computing derivatives. Derivatives are computed by finding the limit of the difference quotient of a function as h approaches 0, like you can see below.

Basically, we can compute the derivative of f(x) using the limit definition of derivatives with the following steps:
1. Find f(x + h).
2. Plug f(x + h), f(x), and h into the limit definition of a derivative.
3. Simplify the difference quotient.
4. Take the limit, as h approaches 0, of the simplified difference quotient.
Example
So consider our racing function f(x) = -7x2 + 280x. First, we find f(x + h):
f(x + h) = -7(x + h)2 + 280(x + h) = -7(x2 + 2xh + h2) + 280x + 280h = -7x2 - 14xh - 7h2 + 280x + 280h
Now, we plug into the limit definition, simplify, and find the limit, as you can see here.

All right. Now that you've done that, we see that the derivative of f(x) is:
f ' (x) = -14x + 280
We can use this formula to calculate the winner's speed at any time during the race. For instance, consider her speed after 10 seconds. We plug x = 10 into the derivative formula:
f ' (x) = -14(10) + 280 = 140

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#### Monoxdifly

A Calculus question in "Pre-Calculus"...
Then define "Pre-Calculus"

#### Prove It

MHF Helper
Then define "Pre-Calculus"
Concepts that are required to be understood before attempting calculus.