# Error computing total derivative

1. Oct 22, 2011

### reemas

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

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

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

however, the correct answer seems to be:
$$3t^2+60t-21$$

Last edited: Oct 23, 2011
2. Oct 22, 2011

### LCKurtz

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

$$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} +\frac{\partial z}{\partial y}\frac{dy}{dt}$$

3. Oct 23, 2011

### 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.

4. Oct 23, 2011

### 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'

Last edited: Oct 23, 2011