Question

implicit differtiating

Answer 1

im sorry @hartnn for being annoying with the differtiating questions

Answer 2

this one is easier than last one
have you tried it ?
2 product rules for 1st 2 terms

Answer 3

but this looks way to different from the ones you helped me with

Answer 4

hey, no problem :)

Answer 5

yeah i tried it and i got this

Answer 6

x^5+5x^4*y+5x*5y^4+5y^5=2

Answer 7

how is first term just x^5 ?
you forgot the chain rule ?

Answer 8

because y is just 1 with you find the derivative of it

Answer 9

right?

Answer 10

oh, but its w.r.t x
so,
d/dx (x^5y ) = x^5 dy/dx + 5x^4 y
got this ?

Answer 11

what do you do when its w.r.t.x?

Answer 12

we use chain rule for 'y'

Answer 13

okay got it

Answer 14

so similarly what you get for term 5xy^5 ?

Answer 15

would it be 5x dy/dx+ 5y^4

Answer 16

diff. of y^5 w.r.t 'x'
will be 5y^4 dy/dx
isn't it ?

Answer 17

why did you add dy/dx?

Answer 18

isnt it already differtiated?

Answer 19

the chain rule!
\(\Large [f(g(x))]' = f'(g(x))\times g'(x)\)

Answer 20

when do you know to use the chain rule? when its like x^5*y?

Answer 21

\(\Large \dfrac{d}{dx}f(y)= f'(y)\dfrac{dy}{dx}\)
another version

Answer 22

we use chain rule whenever we have function of 'x' or 'y'

Answer 23

but sometimes there is x or y and we dont have to use the chain rule...when is it appropriate to use it?

Answer 24

it also applies for ''x but since dx/dx =1, we just ignore it
\(\Large \dfrac{d}{dx}f(x)= f'(x)\dfrac{dx}{dx}=f'(x)\)

Answer 25

for this problem....i dont get how to apply the chain rule..where is f(x) and g(x)?

Answer 26

that is just the general formula...
ok,
for 5xy^5
5 ( y^5 + x (5y^4)dy/dx)

see whether you get this ?

Answer 27

yes i got that but the dy/dx is for the 5y^4 correct?

Answer 28

yes, since y^5 is the function of 'y' , we need to multiply dy/dx to its derivative...

Answer 29

so would the 5y^4 turn into 20y^3?

Answer 30

why would you differentiate 5y^4 again ?
only once was asked,.....

Answer 31

okay got that

Answer 32

so, whats the overall answer ?

Answer 33

so what happens to the x+y on the other side

Answer 34

whats derivative of x ?

Answer 35

1

Answer 36

and derivative of y ?

Answer 37

1

Answer 38

1+!=2?

Answer 39

no!

derivative of y w.r.y x is dy/dx!

Answer 40

ugh i always get confused >.<

Answer 41

so the final answer is 5x^4+x^5*y dy/dx+ 5(y^5-x(5y^4 dy/dx)= 1+dy/dx?

Answer 42

im guessing it incorrect because it looks really long to be an answer

Answer 43

long doesn't mean incorrect....
but,
in 1st term, you missed a 'y'
5x^4y

Answer 44

so the y stays..it doesnt turn into 1 bc of the chain rule?

Answer 45

its because of product rule

Answer 46

okay i see

Answer 47

how come we're not putting dy/dx in fron of the y for 5x^4y?

Answer 48

lets take 1st term and apply product rule (and chain rule)
\( (x^5 y)' = (x^5)'y + x^5 y' \)
got this step ?

Answer 49

now (x^5)' is just 5x^4 , right ?

Answer 50

yes

Answer 51

so now your 1st 2 terms are ?

Answer 52

5x^4y+x^5 y dy/dx?

Answer 53

just
5x^4y+x^5 dy/dx

Answer 54

ok i think i got that part

Answer 55

so the final answer is 5x^4 y +x^5 dy/dx + 5(y^5+x(5y^4 dy/dx)?

Answer 56

=1+dy/dx

Answer 57

similarly for 5xy^5
we get
5y^5 + 25xy^4 dy/dx


and yes, that would be correct final answer.

Answer 58

do i add =1+dy/dx as well?

Answer 59

yup!
\( \Large 5x^4 y +x^5 D+5y^5 + 25xy^4D =1+D \)

Answer 60

but wait! i just realized ,we are not done yet!
we need to find dy/dx
so isolate D from that!

Answer 61

so -1 to both sides?

Answer 62

you need to collect all terms with D on one side....

Answer 63

hmm that seems difficult because there is 2 on the left and one on the other side

Answer 64

i suggest you bring both D's from left to right...

Answer 65

so divide the d from the left?

Answer 66

\(\Large 5x^4 y +5y^5 -1=-x^5 D+D-25xy^4D\)
got this ?
now factor out D from right...

Answer 67

see whether you get
\(\Large D= \dfrac{5x^4y+5y^5-1}{1-x^5-25xy^4}\)