# Homework Help: Do the derivate.

1. Jun 18, 2009

### Penultimate

$$\sqrt{x^2+y^2}$$=2arctg$$\frac{y}{x}$$

2. Jun 18, 2009

### HallsofIvy

Simply saying "Do the derivate" (or derivative) isn't sufficient. The derivatie with respect to what variable? Is the problem to find partial derivatives of that with respect to x and y or are we to assume that y is a function of x and you wish to find dy/dx?

In either case, this is the same as
$$(x^2+ y^2)^{1/2}= 2 arctan(\frac{y}{x})$$
If you want to find dy/dx, can you differentiate $$(x^2+ y^2)^{1/2}$$ with respect to x? What about 2 arctan(y/x)? Of course, you will need to use the chain rule. If f is a function of y, which is a function of x, then df/dx= (df/dy)(dy/dx).

Last edited by a moderator: Jun 19, 2009
3. Jun 18, 2009

### Penultimate

The exercise requires the differential of the expression. I think that is dy/dx.

4. Jun 18, 2009

### qntty

So y is a function of x and the derivative is with respect to x. Attempt the problem first and show where you get stuck.

Recall the chain rule: [f(g(x))]' = f'(g(x)) g'(x). In this case y = g(x)

5. Jun 18, 2009

### Penultimate

I know is something with derivating twice but i am not figuring anything out.

6. Jun 18, 2009

### qntty

Have you learned the chain rule?

7. Jun 19, 2009

### HallsofIvy

So far you haven't shown that you are even trying. What have you done?

8. Jun 19, 2009

### Staff: Mentor

No, the exercise asks for the derivative, dy/dx, not the differential.

9. Jun 19, 2009

### Staff: Mentor

No, you don't need to differentiate twice. Once will be enough to find the derivative, but you will need to use implicit differentiation.