Question

let z=ln(2x^2+y; let x,y be functions of t with x(1)=1, y(1)=2, x'(1)=3 y'(1)=4 find dz/dt when t=1, i know that i start with it in the form
dz/dt=part(z)/part(x)part(x)/part(t) + part(z)/part(x)part(y)/part(t) just not sure where to go from there

Answer 1

find dz/dx and dz/dy
dz/dx = 4x/(2x^2+y)
dz/dy = 1/(2x^2+y)

plug in the given info for when t=1

Answer 2

but how do i find the part(x)/part(t) and part(y)/part(t)? thats the part im stuck on

Answer 3

its given at least for when t=1
x'(1) = 3
y'(1) = 4

Answer 4

thanks, wow not enough coffee today yet i should have seen that, thank you