# Homework Help: Solving for du/dx with multiple variables/equations

1. Aug 14, 2010

### jegues

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

See figure.

2. Relevant equations

N/A

3. The attempt at a solution

Okay, my plan of attack to get du/dx was to solve t in terms of x in one equation, solve s in terms of x in the other and all sub intio the original equation and that the derivative with respect to x.

I can solve t in terms of x,

$$t = \frac{4}{x^3+x^2}$$

However, I'm having trouble doing the same for s. This is about as far as I can get,

$$s = ln\left(s^3x^2 - 1\right)$$

Does anyone have any suggestions?

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2. Aug 14, 2010

### cristo

Staff Emeritus
You don't need to solve the equations. Use implicit differentiation to find dt/dx and ds/dx.

3. Aug 14, 2010

### jegues

Okay so,

$$t = \frac{4}{x^3+x^2}$$

Then,

$$\frac{dt}{dx} = \frac{-12x-8}{x^5 + 2x^4 + x^3}$$

and,

$$\frac{ds}{dx} = \frac{2xs^3}{e^s-3s^2}$$

I'm not sure if I solved for ds/dx or dt/dx correct, but I tried.

Anyone see any problems?

4. Aug 14, 2010

This dt/dx is wrong. You somehow managed to lose a factor of x in every single term. I'm guessing it was a careless mistake.

This ds/dx is correct.

Do you know how to solve for du/dx from here?
$$\frac{du}{dx} = \frac{\partial u}{\partial s}\frac{ds}{dx} + \frac{\partial u}{\partial t}\frac{dt}{dx}$$

5. Aug 14, 2010

### jegues

I actually factored out an x and canceled it... I didnt know I couldn't do that?

6. Aug 14, 2010

You can. Sorry, I guess I just wasn't thinking.

7. Aug 14, 2010

### jegues

Okay since,

$$\frac{du}{dx} = \frac{\partial u}{\partial s}\frac{ds}{dx} + \frac{\partial u}{\partial t}\frac{dt}{dx}$$

Then,

$$\frac{du}{dx} = e^te^s\frac{2xs^3}{e^s-3s^2} + (e^te^s - sint)(\frac{-12x-8}{x^5+2x^4+x^3})$$

8. Aug 14, 2010