Please show your work or process httpi1175photobucketcomalbu

Please show your work or process

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Solution

f(x,y) = x² + y
c(t) = (e^t , e^-t )

I don\'t have the partial symbol available, so I will use d

a)Using the Chain-rule, d f(c(t))/dt = df/dx dx/dt + df/dy dy/dt

df/dx = 2x dx/dt = e^t

df/dy = 1 dy/dt = -e^-t

Thus, d f(c(t))/dt = df/dx dx/dt + df/dy dy/dt = 2x e^t +1 (-e^-t) = (substituting for x)

2e^t e^t -e^-t = 2e^2t  -e^-t

b) Substituting first, we get f(x,y) = x² + y = (e^t)² + e^-t = e^2t + e^-t

Then, df(x,y)/dt = 2e^2t - e^-t

This is the same answer as a)