Can someone please help with the below question?

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dy/dt for ln x =e^y+(x^2+y^3)^2-2 if x=1 y=0  and dx/dt=-2

Jul 15th, 2015

Hi, I am pleased to do this question for you. I am giving the detailed solution below.

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

differentiating this with respect to dt

d(ln x)/dt = d(e^y)/dt + d((x^2+y^3)^2) - d(2)/dt

(1/x)dx/dt = (e^y)dy/dt + 2*((x^2+y^3)(2x*dx/dt + (3*y^2)dy/dt) - 0      - >  (equation 1)


differential of ln x is 1/x 

differential of e^y is e^y but we are doing it with respect to dt so dy/dt term will come

similarly for the next term and differential of constant(2) is always 0.

putting the given values in equation 1,

(1/1)(-2) = (e^0)dy/dt + 2*((1^2 + 0)(2*1*-2 + 0))

-2 = dy/dt + 2*-4

dy/dt = 6      --------->  ANSWER

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 15th, 2015

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