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implicit differentiation: sin((x^2)(y^2))=x

Mathematics
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ok
is this sin [ x^2y^2] = x ? or (sin x^2 ) y^2 = x
first one

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Other answers:

ok
please make me fan
make you fan?
jkwon, let me answer it. i got this
\[\sin(x^2y^2)=x \] \[\cos(x^2y^2)(2xy^2+x^2(2y)(y')=1\] \[2xy^2+x^2(2y)y'=1/[\cos(x^2y^2)]\]
grrrr
oh ok
oh wow
how do you use latex?
Oh I only did half of it
could you explain chain rule?
sure
how do you type latex like type integral x dx
So you treat the problem like an onion. You do the outer layer and then you go to the second layer, then the third. So the outer layer is the cosine, and the inner layer is the x^2y^2
So then....
Oops i mean sine is the outer layer
yeah i gotcha. so you just derive each in parts depending on parenthesis
pretty much yeah ^^
remember the inside is the product
haha, jk i think you have this in the bud
huh?
nvm
well im a math major, so whatever needs help, I like to solve it
whats a non archimedean ordered field?
can you ask that in a separate question? you're distracting him from helping me
so yeah, the product rule for the x^2y^2
yuppp
notgoodatmath, dont speak unless told to
i have 15 fans, i have solved a hundred math problems
what is Y'?
I just want help with this problem, and you aren't helping me. Your're doing the opposite actually.
hey cantorset, use the equation editor in the lower left hand side of the reply box
for latex
oh ok did you see my comment on n^n / n!
oh did not check heh, will do so
i dont see that as an argument what you did to show it converges

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