Question

dy/dx of x^3+y^3=4xy+1

Answer 1

take it term by term,
whats d/dx of x^3 =....?
d/dx y^3 =...?? [Note : here you will need a chain rule]

Answer 2

d/dx of x^3 = 3x^2
is the d/dx of y^3 not just 3x^2 as well? why do i need the chain rule?

Answer 3

because y^3 is NOT the function of 'x'
so, by chain rule,
\(\Large \dfrac{d}{dx}y^3=3y^2\dfrac{dy}{dx}\)
got this ?

Answer 4

ohh, okay, yes, got that. i just didn't realize that was the chain rule, since i didn't see an "outer" or "inner" part.

Answer 5

ok, so we have our left part,
what about the right part ?
d/dx [4xy] =..... ? [Note you will need product rule here.]
d/dx 1 =... ?

Answer 6

d/dx [4xy]= 4(x(dy/dx)+y) which = 4x(dy/dx)+4y ?
d/dx 1 = 0

Answer 7

correct! good :)
so, we'll just need to isolate dy/dx from
\(\Large 3x^2+3y^2\dfrac{dy}{dx}=4x\dfrac{dy}{dx}+4y\)

try first ?

Answer 8

dy/dx (3y^2-4x) = 4y-3x^2 ?

Answer 9

oh my! you are good at this :D
its correct :)
just divide 3y^2-4x and you're done :)

Answer 10

haha, thanks... i understand majority of implicit differentiation, i just need a little help on a few steps here and there.
so my answer would be :
dy/dx=4y-3x^2/3y^2-4x ?
if so, then why does it come out to 3x^2-4y/4x-3y^2 instead on wolfram alpha?

Answer 11

they both are same! (a-b)/(c-d) is same as (b-a)/(d-c)!

Answer 12

both are perfectly valid and accurate answers :)

Answer 13

oh, duh. i remember that now. well thank you so much!!

Answer 14

you're most welcome ^_^
and since you are new here,
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Answer 15

if you have any doubts browsing this site, you can ask me :)

Answer 16

sweet, thanks! if i come up with more questions (which i undoubtedly will), is there a way i can just ask you directly?

Answer 17

you can message me, but a preferable choice will be to post it here,as a new question, so that even if i am not online, you can surely get help from some one else :)