Error computing total derivative

[reemas]

find the total derivative dz/dt, given:
[tex]z=(x^2)-8xy-(y^3)[/tex] where [tex]x=3t[/tex] and [tex]y=1-t[/tex]

my steps look like this, can someone point out where i am going wrong, please?
[tex]z'=2x-8(x'y+y'x)-3y^2[/tex] where
[tex]x'=3[/tex] and [tex]y'=-1[/tex]

[tex]2(3t)-8[3(1-t)+(-1)(3t)]-3(1-t)^2[/tex]
[tex]=6t-8[3-3t+(-3t)]-3(1-2t+t^2)[/tex]
[tex]=6t-8(3-6t)-3(1-2t+t^2)[/tex]
[tex]=6t-24+48t-3+6t-3t^2[/tex]
[tex]=12t+48t-27-3t^2[/tex]
[tex]=-3t^2+60t-27[/tex]

however, the correct answer seems to be:
[tex]3t^2+60t-21[/tex]

i'm going nuts trying to figure out where i went wrong. please help.

 

[LCKurtz]

find the total derivative dz/dt, given:
[tex]z=(x^2)-8xy-(y^3)[/tex] where [tex]x=3t[/tex] and [tex]y=1-t[/tex]

my steps look like this, can someone point out where i am going wrong, please?
[tex]z'=2x-8(x'y+y'x)-3y^2[/tex] where
[tex]x'=3[/tex] and [tex]y'=-1[/tex]

[tex]2(3t)-8[3(1-t)+(-1)(3t)]-3(1-t)^2

=6t-8[3-3t+(-3t)]-3(1-2t+t^2)
=6t-8(3-6t)-3(1-2t+t^2)
=6t-24+48t-3+6t-3t^2
=12t+48t-27-3t^2
=-3t^2+60t-27[/tex]


however, the correct answer seems to be:
[tex]3t^2+60t-21[/tex]

i'm going nuts trying to figure out where i went wrong. please help.

Use just tex, not latex in your tex brackets. You want to use the formula:

[tex]\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt}
+\frac{\partial z}{\partial y}\frac{dy}{dt}
[/tex]

 

[reemas]

thanks for the response, i corrected the formatting. the formula is very helpful.

is there any reason my steps don't work? it seems like I'm applying all the basic principles correctly, so i feel stumped.

 

[lurflurf]

If

z=x²-8x y-y³

then

z'=2x{x'}-8(x' y+x y'))-3y²{y'}

you have omitted the terms inside the {}

or factorize into chain rule form like LCKurtz

z'=(2x-y)x'-(8x+3y²)y'