Question

Find the critical points of f(x) = x^4*(x-1)^3

Answer 1

you first need to find the derivative of the equation
then set the equation to zero
solve for x
and you will have your critical points

Answer 2

\[f(x) =x^{4}(x-1)^{3} \]
is this what your equation looks like?

Answer 3

Yes, it is

Answer 4

I know to do the product rule, but I get stuck after that.

Answer 5

Why did you use the product rule? I was thinking chain rule ? hmm

Answer 6

I end up with: \[x ^{4}3(x-1)^{2}+4x ^{3}(x-1)^{3}\]

Answer 7

Chain rule confuses me quickly. I can understand it with y' notation, but I still can't grasp dy/dx and all of that. I have less than a month left and no more Calculus classes after this.

Answer 8

dy/dx is the same thing as y' (y prime)

Answer 9

I know, but I get lost quickly with the dy over dx and such. Never understood it. I've gotten by with y' notation. Looks cleaner to me

Answer 10

chain rule is simply f'(g(x)) * (g'(x))
so basically
f'=derivative of what's outside multiplied by whats inside (leave the inside alone) and then take derivative of inside...
does that help?

Answer 11

yes y' is better to most folks.

Answer 12

I do get it to an extent. I get lost as to how it applies to this problem.

Answer 13

i like u substitution myself

Answer 14

example

u^n -> n*u^(n-1) * u'

and u ' = du/dx

Answer 15

u substitution captures the 'form' of it

Answer 16

So \[(x-1)^{3}\] becomes \[3(x-1)^{2}* 1\]

Answer 17

3(x-1)^2 * (x-1)'
3(x-1)^2 * 1

Answer 18

Okay, but what happens with the \[x ^{4}\] here?

Answer 19

there you have 4u^3 * u' ,
so here u = x

4(x)^3 * x'
4x^3 * 1

Answer 20

so u substitution works with just x, but you can use the simple power rule when it is just x

Answer 21

Okay

Answer 22

i use the u rule when i have something more complicated than a function with x alone

Answer 23

(x^2 - 3x) ^5 , the derivative of that is hard using only simple power rule. i would have to expand that

Answer 24

I understand. So for my original problem, I still would use the product rule with the chain rule, right?

Answer 25

not the right wording

Answer 26

let us try to recall what the general product rule is
for simplicity I will use the u'v + uv'

which is the same when we split it into two functions f(x) and g(x)
f'(x) * g(x) + f(x) * g'(x)


so we are looking at two derivatives - one for the u and one for the v

in your problem, f(x) = x^4(x-1)^3, in order to get the derivative, you need to identify your u and your v

u = x^4
v = (x-1)^3

now, we obtain their derivatives individually:
u requires only a power-rule 4x^3
v requires the chain-rule 3(x-1)^2 * 1 = 3(x-1)^2

if you're getting confused with chain-rule, think of it as the big fish and the small fish.
big fish being the outermost of the expression without touching the contents inside the parenthesis and the small fish is whatever you have inside the parenthesis. so we obtained the big fish's derivative first then multiply it by the small fish's derivative.

so, we clean things up so you can see what we have so far
PRODUCT RULE
u'v + uv'

u = x^4, u' = 4x^3
v = (x-1)^3, v' = 3(x-1)^2

now apply our product rule