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# C3 differentiation watch

1. 1) f(x) = x^3/2 (x - 3)^3 x > 0

a) show that f ' (x) = k x^1/2 (x - 1)(x - 3)^2 where k is a constant to be found

b) find the coordinates of any stationary points on the curve

i need some help as im not sure how to even properly start this question

i did:

let u = (x-3)^3
y = u^3
du/dx = 1
dy/du = 3U^2
dy/dx = 3(x-3)^2

is this right? and if so where do i go from here?
2. (Original post by sharp357)
1) f(x) = x^3/2 (x - 3)^3 x > 0

a) show that f ' (x) = k x^1/2 (x - 1)(x - 3)^2 where k is a constant to be found

b) find the coordinates of any stationary points on the curve

i need some help as im not sure how to even properly start this question

i did:

let u = (x-3)^3
y = u^3
du/dx = 1
dy/du = 3U^2
dy/dx = 3(x-3)^2

is this right? and if so where do i go from here?
its simpler to use the product rule

you can differentiate x^3/2 and (x - 3)^3 rather easily

(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)
3. ok i'm gettin x^3/2(3(x-3)^2) + 3/2 x^1/2(x-3)^2

but how do i simplify this further
4. (Original post by sharp357)
ok i'm gettin x^3/2(3(x-3)^2) + 3/2 x^1/2(x-3)^2

but how do i simplify this further
well for a start the answer is

=P

Spoiler:
Show

5. (Original post by sharp357)
1) f(x) = x^3/2 (x - 3)^3 x > 0

a) show that f ' (x) = k x^1/2 (x - 1)(x - 3)^2 where k is a constant to be found

b) find the coordinates of any stationary points on the curve

i need some help as im not sure how to even properly start this question

i did:

let u = (x-3)^3
y = u^3
du/dx = 1
dy/du = 3U^2
dy/dx = 3(x-3)^2

is this right? and if so where do i go from here?
f(x) = x^3/2 (x - 3)^3 x > 0
use the product rule:

u= x^3/2 du'= 3/2x^1/2

v= (x-3)^3 dv' l ^3= 3l^2

( l=x-3) l' =1

1* 3l^2= 3(x-3)^2

you then do v*du' + u*dv'

sorry if this is confusing i hope it helps
6. ok but how do i get that to k x^1/2 (x - 1)(x - 3)^2
7. (Original post by sharp357)
ok but how do i get that to k x^1/2 (x - 1)(x - 3)^2
click the spoiler button xD

i've done it all
8. wow thanks for the help

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