help me in finding the derivative of the function below pls
y = 7x^3(x^3-3)^5 / (4x+1)^3(7-x^5)^6

- anonymous

help me in finding the derivative of the function below pls
y = 7x^3(x^3-3)^5 / (4x+1)^3(7-x^5)^6

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- anonymous

as there is a division try applying the quotient rule (vdu-udv/v^2) the numbers are very awkward youll probably want to use a calculator, or derive it with that if your calculator can

- anonymous

entering it how you posted it into mine i got -(60x^16+12x^15-1080x^13-225x^12+7560x^10+1620x^9-25920x^7-5670x^6+43740x^4+9720x^3-29160x-6561)/(16807x^28(4x+1)^4)

- anonymous

@Taplin44 i attached a picture of the function so you can understand clearly. pls help me

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- anonymous

The question asks for dy/dx yes ? if so the verrrry long string of numbers i posted is the correct answer
note that 16807x^28(4x+1)^4 is all the denominator to the rest

- anonymous

@Taplin44 how bout this picture? is it clear for you?

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- amistre64

just pull the bottom up and run a 4 term product rule ...
abcd \(\color{red}{\to}a'bcd+ab'cd+abc'd+abcd'\)

- anonymous

@amistre64 do i have to use quotient rule, product rule, and chain rule?

- amistre64

id just use the product rule on something this messy

- amistre64

\[y=7x^3~(x^3-3)^5~(4x+1)^{-3}~(7-x^5)^{-6}\]

- anonymous

The picture is clear. It is just alot of arithmetic to do by hand hence why i evaluated with a calculator for you
The quotient or product rule will work. Using the product rule requires you to bring up the denominator but as @amistre64 said, is going to be a lot easier

- amistre64

the results will still be a long string tho :) and you would be able to simplify it by factoring out what amounts to the square of the bottom

- anonymous

@taplin44 well our teacher requires us to use the quotient rule, product rule, and chain rule

- amistre64

the quotient rule is pretty well obsolete since dividing is just another way of multiplying.
The rule is a throw back to the limit method i believe

- amistre64

you are still applying the product rule, the power rule, and the chain rule in this setup

- anonymous

okaay thank u :)

- anonymous

@Taplin44 @amistre64 how to find the derivative of the numerator? help

- amistre64

the numerator is just a product ...
ab \(\to\) a'b + ab'

- amistre64

let a = 7x^3
let b=(x^3-3)^5
define their own prospective derivatives and fill in the rule ....

- anonymous

@amistre64 @Taplin44 how to simplify the attached? help

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- anonymous

@rose21 here!

- rose21

yea I see
lol

- anonymous

@rose21 attached is the equation. find the derivative. help me

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- anonymous

you have to use logarithemc diffrentation

- anonymous

@Ahmad1 we haven't encountered that topic yet. all i know is to find the derivative by using quotient rule, product rule, and chain rule

- anonymous

trust me I'm telling you the idea for such questions , it's simple by the way
\[\ln(y)=3\ln(7x)+5\ln(x^3-3)-3\ln(4x+1)-6\ln(7-x^5)\], now differntiate both sides , the LHD is dy/dx *1/y , so multiply by y to get y prime , i.e\[dy/dx=y*(RHS) \prime\] which is easy to find

- anonymous

im sorry but idk what that is

- anonymous

don't you know the logarithmic function ?
I have just used its properties

- anonymous

no sorry. we haven't encountered that yet.

- anonymous

okay no problem ..

- anonymous

@UsArmy3947 here!

- anonymous

@UsArmy3947 find the derivative

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- anonymous

@Ahmad1 how to simplify the attached?

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- anonymous

@rose21 how to simplify the attached?

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- UsArmy3947

ok simplify the attachment on top?

- anonymous

@UsArmy3947 this one

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- UsArmy3947

ok thx

- UsArmy3947

ok i cant help u but ill ask my friend ok?

- anonymous

ok

- anonymous

is this the derivative of the function?

- UsArmy3947

@thomaster @shkrina @robtobey @elisuzsmith @razor99 @terenzreignz @tshark14 @Tron_Cat @YourMentor

- anonymous

@Ashleyisakitty @andriod09 @allie_bear22 @ankit042 @Ashja @ashley_f97

- UsArmy3947

im sorry if i couldn't help u

- anonymous

it's okay :)

- amistre64

its prolly best not to try to "simplify" it; and just to keep it in its "factored" form. at least thats my opinion

- anonymous

@amistre64 u mean that's the final answer?

- UsArmy3947

@amistre64 @Hero @dumbcow @dan815 @dmezzullo @DebbieG @dlapointe25 @Deivyneee @andriod09 @allie_bear22 @ankit042 @Ashleyisakitty @AnElephant @bashirk @iceicebaby

- amistre64

\[y=7x^3~(x^3-3)^5~(4x+1)^{-3}~(7-x^5)^{-6}\]
\[y'=\\
~~~~7(3)x^2~(x^3-3)^5~(4x+1)^{-3}~(7-x^5)^{-6}\\
+35(3)x^5(x^3-3)^4~(4x+1)^{-3}~(7-x^5)^{-6}\\
-21(4)x^3~(x^3-3)^5~(4x+1)^{-4}~(7-x^5)^{-6}\\
+42(5)x^7~(x^3-3)^5~(4x+1)^{-3}~(7-x^5)^{-7}\]
then factoring out a : \((4x+1)^{-6}~(7-x^5)^{-12}\)
\[y'=[(4x+1)^{-6}~(7-x^5)^{-12}]\\
[~~~~7(3)x^2~(x^3-3)^5~(4x+1)^{3}~(7-x^5)^{6}\\
+35(3)x^5(x^3-3)^4~(4x+1)^{3}~(7-x^5)^{6}\\
-21(4)x^3~(x^3-3)^5~(4x+1)^{2}~(7-x^5)^{6}\\
+42(5)x^7~(x^3-3)^5~(4x+1)^{3}~(7-x^5)^{4}]\]

- amistre64

i spose the squaring of the bottom is a bit much ... we could simply factor out the higest negative degrees instead for a more simplified form

- amistre64

\[y'=[(4x+1)^{-4}~(7-x^5)^{-7}]\\
[~~~~7(3)x^2~(x^3-3)^5~(4x+1)~(7-x^5)\\
+35(3)x^5(x^3-3)^4~(4x+1)~(7-x^5)\\
-21(4)x^3~(x^3-3)^5~~(7-x^5)\\
+42(5)x^7~(x^3-3)^5~(4x+1)]\]

- amistre64

factoring out the smallest exponents of the top gets us:
\[y'=[x^2(x^3-3)^4]\\
[~~~~7(3)~(x^3-3)~(4x+1)~(7-x^5)\\
+35(3)x^3~(4x+1)~(7-x^5)\\
-21(4)x~(x^3-3)~~(7-x^5)\\
+42(5)x^5~(x^3-3)~(4x+1)]\\
-----------------\\
~~~~~~~~~~(4x+1)^{4}~(7-x^5)^{7}\]
and theres some constants we could pull as well

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