find dy/dx at x=0 y=u-(2/u) u= (3x+1)^3

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find dy/dx at x=0 y=u-(2/u) u= (3x+1)^3

Mathematics
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dy/du = 1+(2/u^2) du/dx = 3*3(3x+1)^2
2 9(3x+1)^2 (1+ -------- ) * ( ---------) (3x+1)^6 1
18(3x+1)^2 9(3x+1)^2 + ------------ (3x+1)^6 9(3x+1)^6 + 18 --------------- ?? (3x+1)^4

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how did you get dy/du =u-(2/u)?
y = u -(2/u) dy = 1 - (-2/u^2) du dy/du = 1 + (2/u^2)
at x=0; id say that i did the calculation right and go with: 27
ohh got it! thanks!
you could simply plug the value of u into the equation for f(u)... but the chain rule makes life easier

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