# Chain rule: y = f(t, t^2, t^3) and y = g(t, h(t), k(t^2))

1. Mar 30, 2014

### 939

1. The problem statement, all variables and given/known data

I am confused because for each problem there is no equation and for one no intermediate variables.

Compute dy/dt when

a) y = f(t, t^2, t^3)
b) y = g(t, h(t), k(t^2))

2. Relevant equations

a) y = f(t, t^2, t^3)
b) y = g(t, h(t), k(t^2))

3. The attempt at a solution

a)

dy/dt = ∂f/∂t * ∂f/∂t^2 * ∂f/∂t^3

2)

dy/dt = ∂f/∂t + (∂g/∂h * dh/dt) + (∂g/∂k * dk/dt^2)

Last edited: Mar 30, 2014
2. Mar 30, 2014

### LCKurtz

There are intermediate variables, they are just implied so you don't see them. For a) think of $u = t,~v = t^2,~w=t^3$ and $y = f(u,v,w)$. Now do you see how to calculate $\frac{dy}{dt}$ using the chain rule?

3. Mar 30, 2014

### 939

Does this work?

dy/dt = (∂y/∂u * du/dt) + (∂y/∂v * du/dt^2) + (∂y/∂w * dw/dt^3)

4. Mar 30, 2014

### LCKurtz

Look up that chain rule in your calculus book. Is that what it says?

5. Mar 31, 2014

### 939

Sorry... If it was i.e. y = (x, y), x = (t), y = (t^2) I would get it and there my book says you do the same steps as I listed above.

I'm just not sure here... y = f(t, t^2, t^3). Thus, y = dependent. t, t^2, t^3 are independent. Intermediate variables must connect the two... If that is not it, maybe just find derivatives f t, t^2 and t^3 in the equation?

6. Mar 31, 2014

### Staff: Mentor

Yes. Using LCKurtz's suggestion these would be du/dt, dv/dt, and dw/dt.

7. Mar 31, 2014

### LCKurtz

You likely calculated them correctly but I was pointing out your notation was bad. Those two red factors should be dv/dt and dw/dt. The u was probably a typo but you shouldn't have t^2 and t^3 in the formulas.