anonymous
  • anonymous
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
schrodinger
  • 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
  • hartnn
could you implicitly differentiate that ?
hartnn
  • hartnn
you will need chain rule here whats d/dx x^2 =...? whats d/dx y^2=...?
hartnn
  • 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
  • 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
  • hartnn
d/dx (x^2) = 2x thats why and the other 2 was already there
hartnn
  • hartnn
ok, so now what about d/dx (x^2+y^2) =...?
anonymous
  • anonymous
2x+2y?
hartnn
  • 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
  • hartnn
if you got that we are almost done :)
hartnn
  • 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
  • hartnn
whats d/dx x^2 =.... ?
hartnn
  • hartnn
we don't rewrite anything
anonymous
  • anonymous
ohman :/
hartnn
  • 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
  • hartnn
in same way d/dx [(x^2+y^2)^2] = 2 (x^2+y^2) times d/dx (x^2+y^2)
anonymous
  • anonymous
i get that, but in this problem im confused with why the exponent goes away
hartnn
  • hartnn
ok, you said d/dx (x^2) = 2x why did the exponent go away here ?
anonymous
  • anonymous
2(x^2+y^2)^2=25(x^2-y^2)
anonymous
  • anonymous
so you're finding the derivative of the inside?
hartnn
  • hartnn
we do find derivative of inside in chain rule, right ?
anonymous
  • anonymous
yup
anonymous
  • anonymous
my professor also taught us a way of outer first then rewrite inner and then multiple by the derivative of the inner.
hartnn
  • hartnn
he must have meant "derivative of outer first ....."
anonymous
  • anonymous
d/dx[f(g(x))]= f' (g(x))*g'(x)
hartnn
  • hartnn
yes, here f(x) was x2
anonymous
  • anonymous
well isn't it 4?
anonymous
  • anonymous
2(x^2+y^2)^2 |dw:1381774415672:dw|
hartnn
  • hartnn
yes, it is 4.
anonymous
  • anonymous
oh so then it is. where are we getting x2?
hartnn
  • 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
  • hartnn
sorry i didn't get your question?
anonymous
  • anonymous
got that =]
anonymous
  • anonymous
so can it be written like this? : d/dx 4(x^2+y^2) (2x+2y) ?
hartnn
  • hartnn
no, why would d/dx come in front again ?
anonymous
  • anonymous
so i get rid of it?
anonymous
  • anonymous
since i have the derivative already?
hartnn
  • 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
  • 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
  • anonymous
we can't bc we have to leave it as it is right? according to the chain rule
hartnn
  • hartnn
because there, you have not differentiated x^2+y^2 yrt
hartnn
  • hartnn
*yet
hartnn
  • hartnn
and yes, we need to leave it as it is, that too
anonymous
  • anonymous
ok
anonymous
  • anonymous
now what do we do
anonymous
  • anonymous
same for the other side?
hartnn
  • hartnn
yes, other side is too simple. try ?
hartnn
  • hartnn
right side is just 25(2x-2y dy/dx) isn't it ?
anonymous
  • anonymous
25(x^2-y^2)*(2x-2y*dy/dx)
hartnn
  • hartnn
what why ? we need chain rule for just y^2 here, isn't it ?
anonymous
  • anonymous
huh i did?
hartnn
  • hartnn
its like 25x^2 - 25y^2 derivative = 25(2x) - 25 (2y dy/dx)
anonymous
  • anonymous
where did i go wrong?
hartnn
  • hartnn
the x^2-y^2 part was unnecessary, you though it was chain rule again ?
anonymous
  • anonymous
wait it wasn't because it had no exponent?
hartnn
  • hartnn
yes, correct.
anonymous
  • anonymous
25*(2x-2y*dy/dx)
anonymous
  • anonymous
so its this?
anonymous
  • anonymous
got cha.
anonymous
  • anonymous
now do we plug in x and y?
hartnn
  • hartnn
correct! put x=-3, y=1 in that...
hartnn
  • hartnn
i assume you are simplifying...
anonymous
  • anonymous
i don't know i keep getting the answer wrong whenever i try to submit it
hartnn
  • hartnn
what did u get for dy/dx ?
anonymous
  • anonymous
i keep getting incorrects :/
anonymous
  • anonymous
i got 3/119
hartnn
  • 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
  • anonymous
ohi wrote one of the signs wrong.
anonymous
  • anonymous
got it.
anonymous
  • anonymous
thank you so much!
anonymous
  • anonymous
:)
hartnn
  • hartnn
welcome ^_^

Looking for something else?

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