Can someone please help with the below question?

label Calculus
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Thank you for the opportunity to help you with your question!

lnx =e^y +(x^2+y^3)^2-2

dy/dx(lnx ) = dt/dy(e^y +(x^2+y^3)^2-2)

dy/dx= 0 + dy/dx(x^4+y^6+2x^2 y^3)

dy/dx= 4x^3+4x y^2

dy/dx=4+6(1)^2 (0)^2

dy/dx =4

by chain rule

dy/dt = dx/dt * dy/dx

dy/dt = -2 *4

dy/dt = -8

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Jul 14th, 2015

How did you get -8?

My answer is 6...see steps below, did I miss a step? Please advise. Thanks

/x=e^ydy/dx+2(x^2+y^3)(2x+3y^2dy/dx). Put x=1 and y=0: 1=dy/dx+2(1+0)(2+0), so dy/dx=1-4=-3.

dy/dt=-3*-2=6.

Jul 14th, 2015

dear i think my answer is more correct

Jul 15th, 2015

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