Question

Using the chain rule, find the derivative of: (4x+3)^4 * (x+1)^-3

Answer 1

http://www.wolframalpha.com/input/?i=differentiate++%284x%2B3%29^4+*+%28x%2B1%29^-3
Click On Show Steps

Answer 2

fine here...

Answer 3

use quotient rule... then apply u substitution as you go

Answer 4

\[(f * g' - f' * g) / g^2\]

Answer 5

\[(4x+3)^4 * 3(x+1)^3 - 16(4x + 3)^4 * (x+1)^3 / (x+1)^6\]

Answer 6

u get 3(x+1)^3 from..... u = x+1, du = 1

Answer 7

(x+1)^3 then becomes (u)^3... which the derivative is then 3 * (u)^2

Answer 8

er messed up 2 steps up.... it's supposed to be 3(x+1)^2 and 16(4x+3)^3

Answer 9

then substitute the u = x+1 into 3 * (u)^2 and it becomes 3(x+1)^2

Answer 10

do the exact same steps for f' => u = 4x+3, du = 4

Answer 11

then just cancel out the (x+1)^6 with the (x+1)'s in the numerator. you don't have to split the fraction into two but you can

Answer 12

answer then becomes

Answer 13

\[(3(4x+3)^4 * (x+1)^2- 16(4x+3)^3 * (x+1)^3 )/ (x+1)^6\]

Answer 14

\[(3(4x+3)4∗(x+1)2) / (x+1)^6 −(16(4x+3)3∗(x+1)3)/(x+1)^6\]

Answer 15

All i have so far understood is

u = (4x+3)^4
u' = 4(4x+3)^3
v = (x+1)^3
v' = 3(x+1)^2

vu' - uv' / v²

Answer 16

no no no

Answer 17

u = (4x+3) not u = (4x+3)^4

Answer 18

you do this to simplify taking a derivative, so if you were to set u = 4x+3 then the original (4x+3)^4 becomes (u)^4

Answer 19

for the sake of the quotient rule first

Answer 20

we havent even gotten to the chain rule part yet

Answer 21

you did the chain rule with the 4(4x+3)^3

Answer 22

correct me if i'm wrong....
lets say u = (4x+3)^4
u' = 4(4x+3)^3
and lets use....w as the derivative of the inside so w = 4

then for the bottom of the quotient

v = (x+1)^3
v' = 3(x+1)^2
and x for the derivative of the inside so x = 1

Answer 23

now vu'w - uv'x / v²

Answer 24

right however you don't do u substitution in that way

Answer 25

here the idea for this is you can use u and du rather than w or x or whatever

Answer 26

so when you say u = (4x+3)^4... that is an incorrect way of doing it

Answer 27

i understand that but the quotient rule teaches us that vu' - uv' / v²

Answer 28

what you do is you set u = 4x+3 then du = 4 (not w.. du means the derivative of u)

Answer 29

i understand that...there's only so many letters in the alphabet

Answer 30

so then you have (u)^4 * du

Answer 31

er sorry the derivative with respect to u and using the chain rule (u)^3 * du

Answer 32

man this is making no sense

Answer 33

yeah but you have to do u and du... right now it seems trivial but the u and du system will be used in integrals and other stuff which requires the u and du system

Answer 34

START OVER

(4x+3)^4 / (x+1)^3

Answer 35

yeah u substitution is kind of tricky at first

Answer 36

okay

Answer 37

step 1: i wrote the problem
step 2: ?

Answer 38

ok so basically what you do is you use the quotient rule just like you normally would

Answer 39

so uv' - u'v / v^2

Answer 40

k hold on

Answer 41

er u'v - uv' / v^2

Answer 42

(x+1)^3 (4)(4x+3)^3 - 3(x+1)²(4x+3)^4 / (x+1)^6

Answer 43

remember to multiply by du as well.. so on the (4x+3)^4, you brought down the 4 which is 4(4x+3)^3 but you also have to multiply by du which = 4. it becomes 16(4x+3)^3

Answer 44

(4)(x+1)^3 (4)(4x+3)^3 - (1)3(x+1)²(4x+3)^4 / (x+1)^6

???

4 for the derivative of 4x + 3
1 for the derivative of x+1

Answer 45

right 4 for the derivative but you also brought down the 4 from the exponent

Answer 46

(x+1)^3 (16)(4x+3)^3) - 3(x+1)²(4x+3)^4 / (x+1)^6

Answer 47

yeah exactly

Answer 48

u got it

Answer 49

i showed bringing down the 4 from the exponent but hadnt touched the du of the insides yet

Answer 50

now how do you simplify?

Answer 51

k then simplifying is just algebra. so what you can do is split it up into two fractions or you can pull the (x+1)'s out using partial fraction decomposition (fuk this method... way too much work.. use the first)

Answer 52

\[(16(4x+3)^3∗(x+1)^3)/(x+1)^6-(3(4x+3)^4∗(x+1)^2)/(x+1)^6\]

Answer 53

sec let me check this

Answer 54

k yeh its right

Answer 55

so you basically just split up the equation

Answer 56

(16(4x+3)^3∗(x+1)^3)/(x+1)^6 can be (16(4x+3)^3∗(x+1)^3)/ (x+1)^3 (x+1)^3

that splits the denominator into 2 terms of (x+1)^3........... one of those will cancel the (x+1)^3 on top

Answer 57

right so after canceling out you get (x+1)^3 on the bottom

Answer 58

then do the same for the other part

Answer 59

answer in the back of the book:

(4x+3)^3(4x+7) / (x+1)^4

Answer 60

i get 3(4x+3)^4 / (x+1)^4 for the other part

Answer 61

no idea where the (4x+7) term comes from...it is the same answer though

Answer 62

the answer in the back of the book is the same answer. The book used partial fraction decomposition though... which is a pain in the retrice

Answer 63

i think partial fraction was algebra.. dunno you might as well learn it because you will have to use it later in calculus once or twice, but it's really not that useful in my eyes

Answer 64

here check this stuff out on youtube. patrickjmt = calculus god.

Answer 65

chain rule:
http://www.youtube.com/watch?v=6kScLENCXLg

Answer 66

hrm can't find any u substitution from him except for integrals, which you will learn closer to your final. http://www.youtube.com/watch?v=qclrs-1rpKI

Answer 67

ok so i'm starting this other problem now.....this one uses the product rule and makes more sense...

y = sin(x²)cos(2x)

so i'm gonna do uv' d(2x) + vu' d(x²)

Answer 68

they may have not taught you u-sub yet, but i highly recommended getting a good understanding of it early. (it's really not that difficult anyway. u just need a quick vid tutorial on it)

Answer 69

so my answer will be sin(x²)(-sin(2x))(2) + (2x)(cos(2x))(cos(x²)

Answer 70

yeh, just do the same thing. plug and chug. let u = x^2, du = 2x. take derivative of sinu, which = cosu. multiply by du. 2xcos(x^2)

Answer 71

also written as -2sin(x²)sin(2x) + 2xcos(2x)cos(x²)

Answer 72

your answer is right. remember your trig identities though

Answer 73

but it's right

Answer 74

wtf ya'll are doing dot product already? we just finished dot product and i'm in cal 3

Answer 75

dot product?

Answer 76

nm

Answer 77

k got some studying to finish up and you pretty much have it. watch the patrickJMT calculus videos. he's a beast at teaching