Question

how would I differentiate this

Answer 1

\[\frac{ (3x-4)(5x+3) }{ (x-4)^{7} }\]

Answer 2

you may use the product rule and quotient rule or
you can multiply (3x-4)(5x+3) and then use the quotient rule

Answer 3

ahhh but thats wher it becomes confusing as i also have to use the chain rule.

Answer 4

ok lets do it together

(15x^2 -20x +9x-12) / (x-4)^7

(15x^2 -11x - 12) / (x-4)^7

now let
15x^2 -11x - 12 = u
(x-4)^7 = v

using the quotient rule we get

(u'v - v'u) / u^2

so we need to find what are u',v'

u' = 30x - 11
v' = 7(x-4)^6

if you thought about using the chain rule in v' there is no need since there is no "inner" derivative .. for example if you wanted to calculate the derivative of (5x-4)^3
you should use [3*(5x-4)^2] * 5

Answer 5

anyway after i gave you u',v' can you do it ?

Answer 6

im having a look now thanks man. The main purpose of this question is for me to get a better understanding of the chain rule.

Answer 7

but there is no real use of the chain rule here

Answer 8

chain rule says that if you have a function f(g(x)) so df/dx = (df/dg) * (dg/dx)

Answer 9

yea i see what you are showing me with regards to this example but the answer that i get from using chain rule is \[\frac{ (35-21x)(5x+3) }{ (x-4)^{7} }+\frac{ 15x+9 }{ (x-4)^{7} }+\frac{ 15x-25 }{ (x-4)^{7} }\] not your answer of \[\frac{ (30x-11)((x-4)^{7})-(7(x-4)^{6})(15x^{2}-11x-12) }{ ((x-4)^{7})^{2} }\]

Answer 10

obviously the last bit needs to be simplified

Answer 11

@coolsector I think u meant to type v^2 in the denominator in stead typed u^2

Answer 12

ok
(3x-4)(5x+3)

let u = 3x -4 , v = 5x + 3

now [uv]' = u'v+v'u

u' = 3 v' = 5
u'v = 3(5x+3)
v'u = 5(3x-4)

u'v+v'u = 3(5x+3) + 5(3x-4) = 15x + 9 + 15x - 20 = 30x - 11

Answer 13

@ajprincess yes, thank you :)

Answer 14

\[\frac{ 128-54x-75x^{2} }{ x^{8}-32x^{7}+448x^{6}-3504x^{5}+17920x^{4}-57344x^{3}+114688x^{2}-131072x+65536 }\]

Answer 15

so using the quotient rule we get

[ (30x-11) * (x-4)^7 - 7(x-4)^6 * (3x-4)(5x+3) ] / (x-4)^14

Answer 16

yea the simplified version is above

Answer 17

ok

Answer 18

i dont think you should do it..

[ (30x-11) * (x-4)^7 - 7(x-4)^6 * (3x-4)(5x+3) ] / (x-4)^14 =
[ (30x-11) * (x-4) - 7 * (3x-4)(5x+3) ] / (x-4)^8

Answer 19

i think leaving it like this is fine

Answer 20

but my question is how to do this with chain rule not this way. as the answer i need to get to is not the one you supply above but this one \[\frac{ (35-21x)(5x+3) }{ (x-4)^{8} }+\frac{ 15x-9 }{ (x-4)^{7} }+\frac{ 15x-25 }{ (x-4)^{7} }\]

Answer 21

I think its best if we move on to a simpler equation dont you?

Answer 22

say we have \[\frac{ 3x-5 }{ (x-4)^{7} }\] how would i differentiate this using the chain rule?

Answer 23

the first term is the -v'u/v^2
v' = 7*(x-4)^6
u = (3x-4)(5x+3)

so -v'u/v^2 = -(21x-28)(5x+3) / (x-4)^8

this is different from your answer

the second two terms are the u'v/v^2 part which is

(3x-4)5 / (x-4)^7 + (5x+3)4 / (x-4)^7 =

[15x - 20] / (x-4)^7 + [20x+12] / (x-4)^7

Answer 24

so there is something wrong here.. have you posted the question correctly ?

Answer 25

your last question :

u = 3x-5 , v = (x-4)^7

u' = 3 , v' = 7(x-4)^6

[3 / (x-4)^7] + [(35-21x)/(x-4)^8]

Answer 26

wow ok confusion is hurting my brain at the moment

Answer 27

no use of chain rule in here as well

Answer 28

i am now entirely confused by your 2nd last statement

Answer 29

how do you obtain that answer if not by chain rule?

Answer 30

only quotient rule

Answer 31

\[\frac{ f'(x)g(x)-g'(x)f(x) }{ g(x)^{2} }\]????? if you use this you get \[\frac{ 3((x-4)^{7})-(6(x-4)^{7})(3x-5) }{ ((x-4)^{7})^{2} }\] not [3 / (x-4)^7] + [(35-21x)/(x-4)^8]

Answer 32

7(x-4)^6 and not 6(x-4)^7

and its the same

Answer 33

how is that the same? how do you change the exponent from a 7 to an 8 in the denominator?

Answer 34

division

lets take for example the first term
3(x-4)^7 / [(x-4)^7]^2 = 3 / (x-4)^7

Answer 35

yes?

Answer 36

oh lord... my brain has been fried.

Answer 37

so if i have \[\frac{ (x-4)^{6} }{ (x-4)^{14} }\] i can then cancel out the exponents till i have an 8 on the bottom?? I did not know that. And now you have answered my question and my brain has deflated back to normal size once again. That was what I was missing.

Answer 38

\[\frac{ x^{a} }{ x^b } = x^{a-b}\]

Answer 39

i feel like such a knob now.