Question

Implicit function: cos(6xu) = 2u + 7x

Answer 1

a) Calculate the derivative of u with respect to x.
u ' (x) = ??

b) Calculate the derivative of x with respect to u.
x ' (u) = ??

Answer 2

u tried? what u got ?

Answer 3

for u' i had: \[u'=\frac{ -7\sin(6xu) }{ 2 }\]

Answer 4

which def is not right.

Answer 5

cos(6xu) = 2u + 7x
-sin(6xu) (6xu)' = 2u' +7

Answer 6

apply product rule for (6xu)'

Answer 7

would it by 6x * u = (6x)u'+6u

Answer 8

yup. now isolate u'

Answer 9

do i take my 2u' over to the right hand side?

Answer 10

yes

Answer 11

i had: \[u'=\frac{ 2u'+7 }{ -\sin(6xu)(6x) }\]

Answer 12

so then it's u'-2u'= what i just wrote (minus the 2u' on the top)

Answer 13

so it's \[-u'=\frac{ -7 }{ \sin(6xu)(6x) }\]

Answer 14

u didn't isolate u'

Answer 15

i thought i did?

Answer 16

-sin(6xu) ( (6x)u'+6u) = 2u' +7
-sin(6xu) (6x)u' + 6u (-sin(6xu)) -2u' =7
{-sin(6xu) (6x)-2}u'=7- 6u (-sin(6xu))
u' = {7+ 6u (sin(6xu))} / {-sin(6xu) (6x)-2}

just some algebraic manipulations

Answer 17

these are my possible answers, i dont seem to see that one there even though it makes sense to me what you did.

Answer 18

oh wait, those are the answers for part b.
hld one one sec

Answer 19

those are the possible answers for a

Answer 20

those are lots of options....

Answer 21

the second attachment is the list of possible answers.

Answer 22

for a

Answer 23

i can see 4th option matching with our answer.....

Answer 24

{-sin(6xu) (6x)-2}= -2{ 1+3x sin(6xu)}

Answer 25

they just have stuff factored out and they put the 1/2 on the outside of the equation. i see.

Answer 26

did u get all steps ??

Answer 27

yup! i wrote them all down. : ) so now i just do the same thing for part b but, i do it for x' and not u' ?

Answer 28

yeah. u need to diff. again, now w.r.t u

Answer 29

im still getting u' when i do out my derivative.

Answer 30

derivative of u =1

Answer 31

should i write down x as x' and not one?

Answer 32

yup, here x' is dx/du
and not dx/dx

Answer 33

so now i have: \[-\sin(6xu)(6x)(6x')(u)=2+7x'\]

Answer 34

cos(6xu) '
= -sin (6xu) (6xu)'
= -sin (6xu) (6x+6ux')

Answer 35

right, i just wrote it down weird lol.

Answer 36

so that =2+7x'

Answer 37

yup, now algebraic manipulations.....

Answer 38

-sin(6ux)(6ux'+6x)(-sin(6xu)-7x'=2
???

Answer 39

prob not .. lol, this is the part i find the hardest because there's so many variables.

Answer 40

let a=-sin(6ux) (just temporarily)
a(6ux'+6x) = 2+7x
now try

Answer 41

a(6ux'+6x) = 2+7x'

Answer 42

a(6ux')+a(6x)=2+7x'

Answer 43

so... -sin(6ux')-sin(6x)=2+7x'

Answer 44

u still didn't isolate x'

Answer 45

a(6ux')+a(6x)=2+7x'
bring 7x' to left and 6ax to right

Answer 46

then factor x' from left

Answer 47

need help with that ?

Answer 48

x'[a6u-7]=a6x+2

Answer 49

x'[a6u-7]=-a6x+2

Answer 50

thats what i said lol.

Answer 51

so then i sub back in -sin(6ux)

Answer 52

im not seeing my answer on the list of answers tho ..

Answer 53

u just missed a minus....
x'[a6u-7]=-a6x+2
x' = {-a6x+2} / {[a6u-7]}
now put a=-.....

Answer 54

sin(6ux)(6x)+2/-sin(6ux)(6u)-7

Answer 55

2nd option

Answer 56

factor 2 from num., and - from denom.

Answer 57

ohh okY I see!!!

Answer 58

:)

Answer 59

thanks a lot, you're really helpful!

Answer 60

welcome ^_^