Can someone please help with the below question?

label Calculus
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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 11th, 2015

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

dy/dt (ln x )= dy/dt [e^y+(x^2+y^3)^2-2]

dy/dt (ln x )= dy/dt [e^y+(x^4+y^6+ 2 x^2 y^2 )-2]

0 = e^y * dy/dt + y^6 * dy/dt + 2 x^2 y^2 *dy/dt

putting value of x and y

0 = e^0 * dy/dt + 0^6 * dy/dt + 2 *1^2 *0^2 *dy/dt

dy/dt = 0

dx/dt=-2

dy/dx= (dy/dt ) (dx/dt)

dy/dx = 0* -2

dy/dx = 0

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Jul 11th, 2015

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