# Chain Rule for Functions of Two Variables Partial Differentiation Question

1. May 26, 2012

### Kushwoho44

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

Let x=ts^2 -1 and y=ln(s)-t

Use the chain rule for functions of two variables to determine ∂f/∂t at (s,t)=(1,1)

3. The attempt at a solution

y=ln(s)-t

∂f/∂t= ∂f/∂s X ∂s/∂t -1

t=x+1/s^2

∂t/∂s= -2(x+1)/s^3

∂s/∂t=s^3/-2(x+1)

∴ ∂f/∂t= s^2/-2(x+1) -1

= ∂f/∂t= s^2/-2(ts^2) -1

∂f/∂t (1,1) = 1/-2 +1

=0.5

However, the answer is 1 and I have spent two hours trying to work out why I am wrong. All help will be met with great appreciation. Thanks <3

2. May 26, 2012

### HallsofIvy

Staff Emeritus
What is f? Are we to assume that f is some differentiable function of x and y?

$$\frac{\partial f}{\partial t}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$$

$$\frac{\partial f}{\partial s}= \frac{\partial f}{\partial x}(2st)+ \frac{\partial f}{\partial y}\frac{1}{s}$$
At (s, t)= (1, 1), that would be
$$\frac{\partial f}{\partial s}(1, 1)= 2\frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}$$

I have no clue where this is coming from. Before, x and y were functions of s and t. Now t is a function of x and s? Is this a different problem?

Given f a function of x and y and $x=ts^2 -1$ and $y=ln(s)-t$ as before
$$\frac{\partial f}{\partial t}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}= s^2\frac{\partial f}{\partial x}- \frac{\partial f}{\partial y}$$
At (s, t)= (1, 1) that is
$$\frac{\partial f}{\partial t}= \frac{\partial f}{\partial x}- \frac{\partial f}{\partial y}$$

It's impossible to give specific numbers, as you seem to have, because you have not given f as a function of x and y.

3. May 26, 2012

### Kushwoho44

Hi, I'll upload the link to the question. I highly doubt here is an error in the actual question.