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)?

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 our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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)?

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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

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 ?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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 ?
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?
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 ?
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 :/
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 ?
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
my professor also taught us a way of outer first then rewrite inner and then multiple by the derivative of the inner.
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?
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?
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 ^_^

Not the answer you are looking for?

Search for more explanations.

Ask your own question