ChrisV
Find the derivative of the funtion.
f(x)=x^2(x-2)^4
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ChrisV
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\[f(x)=x ^{2}(x-2)^{4}\]
.Sam.
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Use product rule,
y' = (x^2)(4)(x-2)^3+(x-2)^4 (2x)
\[4x^{2}(x-2)^{3}+2x(x-2)^{4}\]
.Sam.
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Teach you one technique,
y=(left)(right)
y'=(copy left)(differentiate right)+(copy right)(differentiate left)
ChrisV
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yea product rule
ChrisV
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you get 2x(x-2)^4+x^2(4(x-2)^3)
ChrisV
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then apply the chain rule which i do not understand
.Sam.
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Noneed to use chain rule here.
ChrisV
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the answer in the back of the book says
ChrisV
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\[x ^{2}[4(x-2)^{3}(1)] + (x-2)^{4}(2x) = 2x(x-2)^{3}(3x-2)\]
ChrisV
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i dont understand hgow to get to 2x(x-2)^3(3x-2)
cuddlepony
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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)
cuddlepony
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easy
cuddlepony
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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
cuddlepony
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After you do this about a 1,000 times you will be able to do it in your head
cuddlepony
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Any questions?
also if you do it in the order I do it you wont mess up using Quotient rule
ChrisV
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sorry let me read what you wrote
ChrisV
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had stepped away
cuddlepony
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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)
cuddlepony
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this is the fool proof way of doing derivatives when you first start out at least it was for me
ChrisV
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ok let me try to reword this
ChrisV
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x^2(x+2)^4
Chlorophyll
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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 ]
ChrisV
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(2x)((x+2)^4) + 4x^2(x-2)^3
cuddlepony
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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)
ChrisV
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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 ?
ChrisV
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and do the product rule again?
cuddlepony
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no!
cuddlepony
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f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2)
That is it you can simplify it but that is the derivative
cuddlepony
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if you wanted to take the second derivative you would use product rule again.
ChrisV
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ok so i need to simplify f'(x) = 2x((x-2)^4) + (4(x-2)^(3))x^(2)
cuddlepony
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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
ChrisV
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so i factor out (x-2)^3
cuddlepony
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you can also factor out 2x
cuddlepony
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2x(x-2)^(3)(x - 2 + 2x)
ChrisV
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so i would end up with 2x(x-2)^3(x-2+2x)
cuddlepony
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2x(3x-2)(x-2)^(3)
cuddlepony
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2x^(1) and x^(1) can be added together
ChrisV
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you have been so much help
ChrisV
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i had the right answer the whole time just didnt see the factor
cuddlepony
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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
cuddlepony
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*patronizing
cuddlepony
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yeah math can be a hassel sometimes lol
ChrisV
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no i understand now
ChrisV
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thanks a lot
cuddlepony
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np take it easy