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ChrisV Group TitleBest ResponseYou've already chosen the best response.0
\[f(x)=x ^{2}(x2)^{4}\]
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.2
Use product rule, y' = (x^2)(4)(x2)^3+(x2)^4 (2x) \[4x^{2}(x2)^{3}+2x(x2)^{4}\]
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.2
Teach you one technique, y=(left)(right) y'=(copy left)(differentiate right)+(copy right)(differentiate left)
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
yea product rule
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
you get 2x(x2)^4+x^2(4(x2)^3)
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
then apply the chain rule which i do not understand
 2 years ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.2
Noneed to use chain rule here.
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
the answer in the back of the book says
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
\[x ^{2}[4(x2)^{3}(1)] + (x2)^{4}(2x) = 2x(x2)^{3}(3x2)\]
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
i dont understand hgow to get to 2x(x2)^3(3x2)
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
f(x)=x^2(x2)^4 split these things up durp, g(x) = x^(2) g'(x) = 2x d(x) = (x2)^4 USE CHAIN RULE!!! s(x) = x^4 s'(x) = 4x^(3) j(x) = x2 j'(x) = 1 so d'(x) =4(x2)^(3)*1 so Now product rule g(x) = x^(2) g'(x) = 2x d(x) = (x2)^4 d'(x) =4(x2)^(3)*1 2x((x2)^4) + (4(x2)^(3))x^(2)
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
f'(x) = 2x((x2)^4) + (4(x2)^(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
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
After you do this about a 1,000 times you will be able to do it in your head
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
Any questions? also if you do it in the order I do it you wont mess up using Quotient rule
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
sorry let me read what you wrote
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
had stepped away
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
Remember Chain rule is just g(x) = x^4 g'(x) = 4x^(3) s(x) = x2 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)
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
this is the fool proof way of doing derivatives when you first start out at least it was for me
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
ok let me try to reword this
 2 years ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
f(x)=x^2(x2)^4 f'(x) = 2x *(x2)^4 + 4x^2 (x 2) ^3 = (x  3) ^3 [ 2x(x2) + 4x^2 ] = (x  3) ^3 [ 6x^2  4x ]
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
(2x)((x+2)^4) + 4x^2(x2)^3
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
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)
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
ah so i then take (2x)((x+2)^4) + 4x^2(x2)^3 and split it like [2x(x+2)^4 +4x^2](x3)^3 ?
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
and do the product rule again?
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
f'(x) = 2x((x2)^4) + (4(x2)^(3))x^(2) That is it you can simplify it but that is the derivative
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
if you wanted to take the second derivative you would use product rule again.
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
ok so i need to simplify f'(x) = 2x((x2)^4) + (4(x2)^(3))x^(2)
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
so i factor out (x2)^3
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
you can also factor out 2x
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
2x(x2)^(3)(x  2 + 2x)
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
so i would end up with 2x(x2)^3(x2+2x)
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
2x(3x2)(x2)^(3)
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
2x^(1) and x^(1) can be added together
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
you have been so much help
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
i had the right answer the whole time just didnt see the factor
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
*patronizing
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
yeah math can be a hassel sometimes lol
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
no i understand now
 2 years ago

ChrisV Group TitleBest ResponseYou've already chosen the best response.0
thanks a lot
 2 years ago

cuddlepony Group TitleBest ResponseYou've already chosen the best response.1
np take it easy
 2 years ago
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