# Very simple question. Yes, I'm stupid.

## Homework Statement

I am trying to solve this basic calculus problem, but I can't figure out the last step. It doesn't make sense to me. I know I'm just being an idiot and not thinking of it. This actually has little to do with calculus and more to do with basic algebra. Doh!

## Homework Equations

The problems asks us to find the critical points of the function f(x) = x^3/5 * (4-x).

## The Attempt at a Solution

I've gotten down to this point:
f'(x) = (3(4-x) / 5x^2/5) - x^2/5.

However, the solution to the problem goes on to this:
f'(x) = (3(4-x) - 5x) / 5x^2/5

I cannot for the life of me see how they reduced it down to that answer.

Any help would be much appreciated!!!

I've gotten down to this point:
f'(x) = (3(4-x) / 5x^2/5) - x^2/5.
The highlighted 2 should be a 3. Then you can factor out the x2/5

Mark44
Mentor
Given that $$f(x) = (4 - x) x^\frac{3}{5}$$
it's simpler to just multiply this out before taking the derivative.
$$f(x) = 4x^\frac{3}{5} - x^\frac{8}{5}$$
$$f'(x) = \frac{12}{5}x^\frac{-2}{5} - \frac{8}{5}x^\frac{3}{5}$$
$$= x^\frac{-2}{5} (\frac{12}{5} - \frac{8}{5}x)$$
Now it's easy to find the critical point.

I didn't think you could multiply it out?

Don't you need to use the product rule to differentiate with respect to x?

Also, there was as typo above. The x^2/5 ought to have been x^3/5.

You can always simplify an expression before taking the derivative if this makes the calculation shorter... note that the derivative of x3/5 is 3/5 x-2/5

I'm obviously rusty on my algebra here, but I greatly appreciate the help you guys are providing. All of it has really helped push me in the right direction. Also, I didn't know you could simplify before differentiating like that. Thank you for showing me that.

I can see it if I simplify before hand; however, the professor is asking us to use the product rule. When I use the product rule, I can't see it, but I am getting closer.

So here is where I'm at:

I went from (3/5x^-2/5(4-x)) - x^3/5

to 3/5x^-2/5((4-x) - 5/3x).

This is close to the answer the book has of 3/5x^-2/5((4-x) - 5x).

My question is, how did 5/3x turn into 5x?

Mark44
Mentor
I'm obviously rusty on my algebra here, but I greatly appreciate the help you guys are providing. All of it has really helped push me in the right direction. Also, I didn't know you could simplify before differentiating like that. Thank you for showing me that.

I can see it if I simplify before hand; however, the professor is asking us to use the product rule. When I use the product rule, I can't see it, but I am getting closer.

So here is where I'm at:

I went from (3/5x^-2/5(4-x)) - x^3/5

to 3/5x^-2/5((4-x) - 5/3x).

This is close to the answer the book has of 3/5x^-2/5((4-x) - 5x).

My question is, how did 5/3x turn into 5x?

Please help us out here. If you're going to write expressions on a single line, use parentheses liberally. It's really hard to tell when you write -x^3/5 whether you are talking about $$\frac{-x^3}{5}$$ or $$-x^\frac{3}{5}$$, and that's a simple example.

For the example I used, write it as -x^(3/5)

I can see where the confusion is. My apologies.

So here is where I'm at, making liberal use of parenthesis:

I went from (((3/5)*(x^(-2/5)))*(4-x)) - (x^(3/5))

to ((3/5)*(x^(-2/5)))*((4-x) - ((5/3)*x))

This is close to the answer the book has of ((3/5)*(x^(-2/5)))*((4-x) - 5x).

My question is, where did I go wrong by using ((5/3)*x), since the book has -5x? Or rather, how do I get to -5x from where I'm at?

If there is any further confusion, please let me know, and I will try to clarify as best as possible.

Mark44
Mentor
I am interpreting the answer you said in your previous post was the book's in this way:
$$\frac{3}{5} x^ \frac{-2}{5} (4 - x - 5x)$$
This is not the right answer, and is in disagreement with what you said was the book's answer in your first post in this thread, namely
f'(x) = (3(4-x) - 5x) / 5x^(2/5) (I added parentheses around the exponent.)

This can be simplified to (12 - 3x - 5x)/(5x^(2/5)) = (12 - 8x)/(5x^(2/5))

This is the same as (4/5)(3 - 2x)/(x^(2/5)), and is equal to what I got using two techniques.

1. By multiplying first and then differentiating, I got
$$f'(x) = \frac{4}{5} x^ \frac{-2}{5} (3 - 2x)$$

2. By using the product rule and then pulling out a common factor, I got
$$f'(x) = \frac{3}{5} x^ \frac{-2}{5} (4 - x ) - x^\frac{3}{5}$$
$$= x^\frac{-2}{5} (\frac{3}{5} (4 - x ) - x)$$
$$= x^\frac{-2}{5} (\frac{12}{5} - \frac{3}{5}x - x)$$
$$= x^\frac{-2}{5} (\frac{12}{5} - \frac{8}{5}x)$$
$$= \frac{4}{5}x^\frac{-2}{5} (3 - 2x})$$

Either of my answers could also be written as
$$f'(x) = \frac{8}{5}x^\frac{-2}{5} (1.5 - x})$$

The two expression I ended with in each case is preferred since you can do something with it, namely find where f'(x) = 0.

Since I got the same answer using two different techniques, I'm pretty confident that I have the right answer. If your answer and/or the books answer can't be converted to look like mine, one or both of the answers you show are incorrect.

Yes, I see what I am looking for. I was making a very small and simple mistake in the factoring, but it obviously had big consequences for my answer.

I am not that particularly great at factoring, so this has really helped me.

I am working to get better, and you guys have helped me progress. Thank you.