Question

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

Answer 1

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

Answer 2

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)

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

just pull the bottom up and run a 4 term product rule ...

abcd \(\color{red}{\to}a'bcd+ab'cd+abc'd+abcd'\)

Answer 7

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

Answer 8

id just use the product rule on something this messy

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

okaay thank u :)

Answer 16

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

Answer 17

the numerator is just a product ...

ab \(\to\) a'b + ab'

Answer 18

let a = 7x^3
let b=(x^3-3)^5

define their own prospective derivatives and fill in the rule ....

Answer 19

@amistre64 @Taplin44 how to simplify the attached? help

Answer 20

@rose21 here!

Answer 21

yea I see
lol

Answer 22

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

Answer 23

you have to use logarithemc diffrentation

Answer 24

@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

Answer 25

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

Answer 26

im sorry but idk what that is

Answer 27

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

Answer 28

no sorry. we haven't encountered that yet.

Answer 29

okay no problem ..

Answer 30

@UsArmy3947 here!

Answer 31

@UsArmy3947 find the derivative

Answer 32

@Ahmad1 how to simplify the attached?

Answer 33

@rose21 how to simplify the attached?

Answer 34

ok simplify the attachment on top?

Answer 35

@UsArmy3947 this one

Answer 36

ok thx

Answer 37

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

Answer 38

ok

Answer 39

is this the derivative of the function?

Answer 40

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

Answer 41

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

Answer 42

im sorry if i couldn't help u

Answer 43

it's okay :)

Answer 44

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

Answer 45

@amistre64 u mean that's the final answer?

Answer 46

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

Answer 47

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

then factoring out a : \((4x+1)^{-6}~(7-x^5)^{-12}\)

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

Answer 48

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

Answer 49

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

Answer 50

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