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could you implicitly differentiate that ?

you will need chain rule here
whats d/dx x^2 =...?
whats d/dx y^2=...?

what ? where does 2*2 come from ?

d/dx (x^2) = 2x
thats why
and the other 2 was already there

ok, so now what about d/dx (x^2+y^2) =...?

2x+2y?

if you got that we are almost done :)

d/dx[2(x^2+y^2)^2]
= 2*(2 (x^2+y^2)) times d/dx(x^2+y^2)
this part ?

whats d/dx x^2 =.... ?

we don't rewrite anything

ohman :/

in same way
d/dx [(x^2+y^2)^2] = 2 (x^2+y^2) times d/dx (x^2+y^2)

i get that, but in this problem im confused with why the exponent goes away

ok, you said d/dx (x^2) = 2x
why did the exponent go away here ?

2(x^2+y^2)^2=25(x^2-y^2)

so you're finding the derivative of the inside?

we do find derivative of inside in chain rule, right ?

yup

he must have meant "derivative of outer first ....."

d/dx[f(g(x))]= f' (g(x))*g'(x)

yes,
here f(x) was x2

well isn't it 4?

2(x^2+y^2)^2 |dw:1381774415672:dw|

yes, it is 4.

oh so then it is. where are we getting x2?

d/dx[2(x^2+y^2)^2]
= 2*(2 (x^2+y^2)) times d/dx(x^2+y^2) = 4(x^2+y^2)times d/dx(x^2+y^2)

sorry i didn't get your question?

got that =]

so can it be written like this? : d/dx 4(x^2+y^2) (2x+2y) ?

no, why would d/dx come in front again ?

so i get rid of it?

since i have the derivative already?

yeah i got it, but why don't we do dy/dx here: 4 (x^2+y^2) times (2x+2y dy/dx)|dw:1381774790286:dw|

we can't bc we have to leave it as it is right? according to the chain rule

because there, you have not differentiated x^2+y^2 yrt

*yet

and yes, we need to leave it as it is, that too

ok

now what do we do

same for the other side?

yes, other side is too simple. try ?

right side is just 25(2x-2y dy/dx)
isn't it ?

25(x^2-y^2)*(2x-2y*dy/dx)

what why ?
we need chain rule for just y^2 here, isn't it ?

huh i did?

its like 25x^2 - 25y^2
derivative = 25(2x) - 25 (2y dy/dx)

where did i go wrong?

the x^2-y^2 part was unnecessary, you though it was chain rule again ?

wait it wasn't because it had no exponent?

yes, correct.

25*(2x-2y*dy/dx)

so its this?

got cha.

now do we plug in x and y?

correct!
put x=-3, y=1 in that...

i assume you are simplifying...

i don't know i keep getting the answer wrong whenever i try to submit it

what did u get for dy/dx ?

i keep getting incorrects :/

i got 3/119

4 (x^2+y^2) times (2x+2y dy/dx) = 25(2x-2y dy/dx)
x=-3, y=1
4(9+1) (-6 +2 dy/dx) = 25 (-6 -2 dy/dx)

ohi wrote one of the signs wrong.

got it.

thank you so much!

:)

welcome ^_^