Derivative of sqrt(x^2+y^2)=2arctg(y/x)

[Penultimate]

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

Dont know where to start with this one either.

 

[HallsofIvy]

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

Dont know where to start with this one either.

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).

 

[Penultimate]

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

 

[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)

 

[Penultimate]

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

 

[qntty]

Have you learned the chain rule?

 

[HallsofIvy]

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

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

 

[Mark44]

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

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

 

[Mark44]

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

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