Find the slope of the tangent line to the curve (a lemniscate) 2(x^2+y^2)^2=25(x^2-y^2) at the point (-3,1)?

- anonymous

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- hartnn

could you implicitly differentiate that ?

- hartnn

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

- hartnn

what ? where does 2*2 come from ?

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## More answers

- hartnn

d/dx[2(x^2+y^2)^2]
= 2*(2 (x^2+y^2)) times d/dx(x^2+y^2)
this is because of chain rule, got this step ?

- hartnn

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

- hartnn

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

- anonymous

2x+2y?

- hartnn

here, chain rule again comes into picture
d/dx f(y) = f'(y) dy/dx
so d/dx (y^2) = 2y dy/dx
got this ?

- hartnn

if you got that we are almost done :)

- hartnn

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

- hartnn

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

- hartnn

we don't rewrite anything

- anonymous

ohman :/

- hartnn

derivative of outer times derivative of inner
like for (sin x)^3
we have outer = x^3, derivative = 3x^2
inner = sin x derivative = cos x
so,
3 (sin x)^2 cos x
got this ?

- hartnn

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

- anonymous

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

- hartnn

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

- anonymous

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

- anonymous

so you're finding the derivative of the inside?

- hartnn

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

- anonymous

yup

- anonymous

my professor also taught us a way of outer first then rewrite inner and then multiple by the derivative of the inner.

- hartnn

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

- anonymous

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

- hartnn

yes,
here f(x) was x2

- anonymous

well isn't it 4?

- anonymous

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

- hartnn

yes, it is 4.

- anonymous

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

- hartnn

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)

- hartnn

sorry i didn't get your question?

- anonymous

got that =]

- anonymous

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

- hartnn

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

- anonymous

so i get rid of it?

- anonymous

since i have the derivative already?

- hartnn

d/dx [(x^2+y^2)^2] = 2 (x^2+y^2) times d/dx (x^2+y^2)
= 4 (x^2+y^2) times (2x+2y dy/dx)
see whether you get this?

- anonymous

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|

- anonymous

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

- hartnn

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

- hartnn

*yet

- hartnn

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

- anonymous

ok

- anonymous

now what do we do

- anonymous

same for the other side?

- hartnn

yes, other side is too simple. try ?

- hartnn

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

- anonymous

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

- hartnn

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

- anonymous

huh i did?

- hartnn

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

- anonymous

where did i go wrong?

- hartnn

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

- anonymous

wait it wasn't because it had no exponent?

- hartnn

yes, correct.

- anonymous

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

- anonymous

so its this?

- anonymous

got cha.

- anonymous

now do we plug in x and y?

- hartnn

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

- hartnn

i assume you are simplifying...

- anonymous

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

- hartnn

what did u get for dy/dx ?

- anonymous

i keep getting incorrects :/

- anonymous

i got 3/119

- hartnn

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)

- anonymous

ohi wrote one of the signs wrong.

- anonymous

got it.

- anonymous

thank you so much!

- anonymous

:)

- hartnn

welcome ^_^

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