Why add a second term when differentiating y=2x(1-x)^2

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
5 replies · 2K views
fitz_calc
Messages
41
Reaction score
0

Homework Statement


y=2x(1-x)^2


Homework Equations





The Attempt at a Solution



y=2x(1-x)^2

y`=-x
-----------------
(1-x)^1/2

I thought i was done here. The book takes a few more steps. it adds:

-x
-----------------
(1-x)^1/2

to this:

2(1-x)^1/2 (1-x)^1/2
------------ x -----------
1 (1-x)^1/2


Why does the last step exist?
 

Discussion

Physics news on Phys.org
and one more question. i just got to a problem with sqrt(x+1) in the denominator. my original thought was to put it in the numerator to the power of -1/2 and use the product rule.

my book uses the quotient rule instead. how do i know which one is the best one to use?
 
Last edited:
fitz_calc said:

Homework Statement


y=2x(1-x)^2

Homework Equations


The Attempt at a Solution



y=2x(1-x)^2

y`=-x
-----------------
(1-x)^1/2

I thought i was done here. The book takes a few more steps. it adds:

-x
-----------------
(1-x)^1/2

to this:

2(1-x)^1/2 (1-x)^1/2
------------ x -----------
1 (1-x)^1/2Why does the last step exist?

Are the dotted lines supposed to indicate that ir is a fraction?

And have you used the chain rule yet?

What is the original function now?
Is it [tex]y=2x(1-x)^2[/tex] ?
 
Last edited:
Did you do the product rule right?

f'(x)g(x) + f(x)g'(x)?

Or secondly, you can follow the definition of the limit and right

lim as x approaches deltax 2(x+dx)(1-(x+dx)^2 all over dx

where dx = delta x

Use that to confirm if yo did the product rule right, I can't understand the way you wrote your work. Looking at your original function, I don't understand how you got a fraction.
 
Last edited:
PowerIso said:
Did you do the product rule right?

f'(x)g(x) + f(x)g'(x)?

PowerIso, can you understand what he's writing? If the original function is just [tex]y=2x(1-x)^2[/tex]

I see no need for the product rule as this is just a polynomial of the third degree...Power Rule.

Casey

Edit: I saw your edit...the fraction is messing me up too..that is why I believe he maybe copied the original function wrong...maybe he meant something else.:confused:
 
Last edited:
Yes I can understand that, but his question regarded the product rule, so I tackled this problem with that view.