Question

Find the derivative of the funtion.
f(x)=x^2(x-2)^4

Answer 1

\[f(x)=x ^{2}(x-2)^{4}\]

Answer 2

Use product rule,

y' = (x^2)(4)(x-2)^3+(x-2)^4 (2x)
\[4x^{2}(x-2)^{3}+2x(x-2)^{4}\]

Answer 3

Teach you one technique,
y=(left)(right)
y'=(copy left)(differentiate right)+(copy right)(differentiate left)

Answer 4

yea product rule

Answer 5

you get 2x(x-2)^4+x^2(4(x-2)^3)

Answer 6

then apply the chain rule which i do not understand

Answer 7

Noneed to use chain rule here.

Answer 8

the answer in the back of the book says

Answer 9

\[x ^{2}[4(x-2)^{3}(1)] + (x-2)^{4}(2x) = 2x(x-2)^{3}(3x-2)\]

Answer 10

i dont understand hgow to get to 2x(x-2)^3(3x-2)

Answer 11

f(x)=x^2(x-2)^4

split these things up durp,

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

d(x) = (x-2)^4
USE CHAIN RULE!!!

s(x) = x^4
s'(x) = 4x^(3)
j(x) = x-2
j'(x) = 1

so
d'(x) =4(x-2)^(3)*1

so Now product rule
g(x) = x^(2)
g'(x) = 2x

d(x) = (x-2)^4
d'(x) =4(x-2)^(3)*1

2x((x-2)^4) + (4(x-2)^(3))x^(2)

Answer 12

easy

Answer 13

f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2)

Best way to deal with derivatives when you are first learning them is to split them into different functions and derive them then put them all together

Answer 14

After you do this about a 1,000 times you will be able to do it in your head

Answer 15

Any questions?
also if you do it in the order I do it you wont mess up using Quotient rule

Answer 16

sorry let me read what you wrote

Answer 17

had stepped away

Answer 18

Remember Chain rule is just
g(x) = x^4
g'(x) = 4x^(3)
s(x) = x-2
s'(x) = 1

we throw away g(x) and sub s(x) into g'(x) and multiply by s(x) derivative
so g'(s(x)) * s'(x)

Answer 19

this is the fool proof way of doing derivatives when you first start out at least it was for me

Answer 20

ok let me try to reword this

Answer 21

x^2(x+2)^4

Answer 22

f(x)=x^2(x-2)^4
f'(x) = 2x *(x-2)^4 + 4x^2 (x -2) ^3
= (x - 3) ^3 [ 2x(x-2) + 4x^2 ]
= (x - 3) ^3 [ 6x^2 - 4x ]

Answer 23

(2x)((x+2)^4) + 4x^2(x-2)^3

Answer 24

g(x)d(x)

Take derivative of
g(x)

take derivative of
d(x) you have to use chain rule to do so

g'(x)d(x) + g(x)d'(x)

if you do it in this order it works for quient rule as well

g(x)/d(x)

(g'(x)(d(x)) - g(x)d'(x))/(d(x))^(2)

Answer 25

ah so i then take (2x)((x+2)^4) + 4x^2(x-2)^3 and split it like [2x(x+2)^4 +4x^2](x-3)^3 ?

Answer 26

and do the product rule again?

Answer 27

no!

Answer 28

f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2)
That is it you can simplify it but that is the derivative

Answer 29

if you wanted to take the second derivative you would use product rule again.

Answer 30

ok so i need to simplify f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2)

Answer 31

its only asking for the first derivative so yeah. Is there anything you do not understand about the method I showed you? If you have to you can

Answer 32

so i factor out (x-2)^3

Answer 33

you can also factor out 2x

Answer 34

2x(x-2)^(3)(x - 2 + 2x)

Answer 35

so i would end up with 2x(x-2)^3(x-2+2x)

Answer 36

2x(3x-2)(x-2)^(3)

Answer 37

2x^(1) and x^(1) can be added together

Answer 38

you have been so much help

Answer 39

i had the right answer the whole time just didnt see the factor

Answer 40

because they are to the same power, remember they have to be to the same power to add them you can't add x^(2) + x^(1) or x^(3) + x

Not to be partrionizing

Answer 41

*patronizing

Answer 42

yeah math can be a hassel sometimes lol

Answer 43

no i understand now

Answer 44

thanks a lot

Answer 45

np take it easy