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