# Do the derivate.

1. Jun 18, 2009

### Penultimate

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

2. Jun 18, 2009

### HallsofIvy

Staff Emeritus
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: 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

Staff Emeritus
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.