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

- schrodinger

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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 ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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 ^_^

Looking for something else?

Not the answer you are looking for? Search for more explanations.