Implicit differentiation: find y'(1)

Limits, differentiation, related rates, integration, trig integrals, etc.
Coolmpl2
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Implicit differentiation: find y'(1)

Postby Coolmpl2 » Thu Oct 11, 2012 12:11 am

If 3x^2+2xy+y^2=2 then what is the value of y'(1)?
My steps so far:
6x+2xy'+2y+2yy'=0
3x+xy'+y+yy'=0
Yy'+xy'=-3x-y
Y'=(-3x-y)/(y+x)

Would it be possible to find this or would the answer not be defined due to the two variables?

nona.m.nona
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Re: Implicit differentiation: find y'(1)

Postby nona.m.nona » Thu Oct 11, 2012 1:21 pm

Coolmpl2 wrote:If 3x^2+2xy+y^2=2 then what is the value of y'(1)?

When x = 1, what is the correspondent value of y, based on the above equation? Use this to complete the computations.


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