Question

Question related to derivatives

A curve is defined by y=x^2(x-2)^2

Show that dy/dx =4x(x-1)(x-2)

Find the coordinated of the stationary points on the curve

Observe the change in the sign of dy/dx as x increases through each of the points and hence deduce the nature of the points.


Please and thank u!!:)

Answer 1

Do you need help with the entire question? I assume you are aware of derivative rules?

Answer 2

first expand the given eqution y=x^2(x^2+4-4x)
y=x^4+4x^2-4x^3
then appliy the dirivation.....

Answer 3

for the first part just split f(x) into smaller functions take their derivatives then sub them into the product rule, which is:

f'(x) = s'(x)g(x) + s(x)g'(x)

Answer 4

chetan552 I wouldn't expand

Answer 5

He just needs to apply chain rule

Answer 6

yes...@Australopithecus but with remebering the formula we can do this process...

Answer 7

chetan552 there is nothing wrong with your method I just feel chain rule is faster and it is always good practice

Answer 8

Kristy18 do you know how to take the derivative of this or do you want me to show you?

Answer 9

your approch is good...@Australopithecus

Answer 10

Why are people convinced that this site is meant to help them cheat?

Answer 11

Well I dont think it is fair to assume that is the case just yet thatguyintheback

Answer 12

Excuse me! I am not cheating! I am trying to do my work at home on my own! I am HOMESCHOOLED!!!!!!!

Answer 13

Dont get defensive kristy18

Answer 14

I know how to take the derivatives but the answer kept coming out incorrect.

Answer 15

we are here to help you tell me what exactly you are having problems with in this question

Answer 16

want me to walk you through it

Answer 17

Sorry lol Just dont like to be accused of cheating

Answer 18

or you can show me your steps, well he is probably just frustrated

Answer 19

yeah sure that would be great thanks

Answer 20

Well as I stated first can you split it into smaller functions?

Answer 21

show me the two functions that y=x^2(x-2)^2 can be split into

Answer 22

yup I did a prev chapter on differentiation chain rules and all that

Answer 23

ummm x^2
and (x-2)^2 (-2)

Answer 24

so you have,

s(x) = x^2

and

g(x) = (x-2)^2

I dont know where that negative 2 came from but I will pretend you didnt write that because it isnt in f(x)

Answer 25

ok

Answer 26

I apologize, Kristy, Calc is a pain to me too.

Answer 27

no problem and yeah I agree!

Answer 28

Now take the derivative of s(x)

before you take the derivative of g(x) you need to split it into smaller functions, can you show me what it splits into?

Answer 29

what do u mean?

Answer 30

With derivatives you always want to split the function into smaller functions and make them easier to work with

Answer 31

i guess it would be 2(x-2)

Answer 32

so I will show you how to do it with g(x) because it isn't obvious

Answer 33

g(x) = (x-2)^2

we need to apply chain rule so we need to split g(x) into smaller functions,

so we have,

m(x) = x^2

l(x) = x-2

Answer 34

Trust me this method is the best way to take derivatives you will eventually just be able to look at something and come up with the derivative once you practice enough

Answer 35

Oh right! u used the factorisation rules!

Answer 36

d/dx(f1(f2(a)))=f2(a)f1'(f2(a))), right?

Answer 37

remember chain rule is

m'(l(x))*l'(x) = g'(x)

Answer 38

so show me the derivatives of the functions I asked you for and I will help you put it all together

Answer 39

I will also put all the work into one post so you can see how derivatives are done

Answer 40

Darn, I garbled it.

Answer 41

are you following kristy?

Answer 42

Help help help

Answer 43

yeah just writing it out thanks just give me twi secs

Answer 44

by l'(x) you mean 1/(x-2)^2 right?

Answer 45

from,

f(x)=x^2(x-2)^2

we split it into two functions

s(x) = x^2
s'(x) = 2x

g(x) = (x-2)^2
g'(x) = ??

we need to apply chain rule so we need to split g(x) into smaller functions, so we have,

m(x) = x^2
m'(x) = 2x

l(x) = x-2
l'(x) = 1

sub into chain rule
m'(l(x))*l'(x) = g'(x)

Now just by looking at the original function,
f(x)=x^2(x-2)^2
you can tell you will need to use product rule, which is,

f'(x) = s'(x)g(x) + s(x)g'(x)

Answer 46

notice that m(x) is the outer part of the function, and l(x) is the inner part of the function, (x-2)^2

Answer 47

Omg! Ok I have missed some stuff!

Answer 48

I'm kind of annoyed that this isnt the method taught normally

Answer 49

derivatives are so easy once you know this method

Answer 50

ok well just to clarify I havent actually done this in school

Also what level is this?

Answer 51

very funy i need to join the students

Answer 52

I am studying Cambridge Olevel Additional Math

Answer 53

this is just how you take derivatives Kristy18

Answer 54

you know what f(x) and g(x) is right?

Answer 55

the notation?

Answer 56

Oh ok just went over it its making more sense now thanks!

Answer 57

I havent been writing my answers in this form

Answer 58

this is highschool level

Answer 59

that notation is very useful

Answer 60

it is worth using, because it lets you talk about multiple functions with ease and specific points in functions

Answer 61

ok cool so I need to learn it now!

Answer 62

it helps you keep track of stuff to when you are doing derivatives

Answer 63

f(x) is the same as y, it is just another way to write y

Answer 64

ok

Answer 65

do you understand how to apply the chain rule formula,

m'(l(x))*l'(x) = g'(x)

Answer 66

in my book they write it as dy/dt=dy/dx times dx/dt

Answer 67

ok so how do you do this part? Observe the change in the sign of dy/dx as x increases through each of the points and hence deduce the nature of the points.

Answer 68

well this is how I do my derivatives, and I never get a wrong answer

Answer 69

The first question is asking you were dy/dx = 0

Answer 70

Ok thanks I will look at it again

Answer 71

at least that is what I gather from that question

Answer 72

What I do is make a table for these questions

Answer 73

yeah when I looked at the examples dy/dx always seemed to equal 0.

Observe the change in the sign of dy/dx as x increases through each of the points and hence deduce the nature of the points.
This is the part which was confusing me the most it seems like a new concept

Answer 74

this is the table I make

Answer 75

so you write down,

set f'(x) = 0

then you find the values of x that give you 0

Answer 76

what is this table for?

Answer 77

Once you find the values of x that give you 0 for the function write them on the table,

so just say my function has x = 2 as a root of f'(x)

I would add it to the table,



if I had x = 3 also as a root of f'(x)