Derivative of a function with ln

1. Oct 16, 2007

ns5032

This is a small part of a bigger problem, but the part I am having trouble with is finding the derivative of...
ln[sqrt(x^2+y^2)]
I'm sure it is something simple and I remember learning it in Calc I or II but I forgot. Please help remind me! Thank you!!

2. Oct 16, 2007

rock.freak667

$$ln(\sqrt{x^2+y^2}) = ln(\sqrt{x^2+y^2})^\frac{1}{2} = \frac{1}{2}ln(x^2+y^2)$$

Is it easier now?

3. Oct 16, 2007

ns5032

Yes I believe so...

Would it then be...

(1/2) [(2x+2y)/(x^2+y^2)]

...if I remember correctly??

4. Oct 16, 2007

ns5032

Which would then simplify to

1/(x+y)

?

5. Oct 16, 2007

supraanimo

Are you taking the total derivative, or what is the variable you are differentiating with respect to?

6. Oct 16, 2007

rock.freak667

Remember $$\frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx}$$

7. Oct 17, 2007

HallsofIvy

Staff Emeritus
Supraanimo's question is still important. ln[sqrt(x^2+y^2)] has two variables. "The derivative" might be the gradient, but more often you are asked to find the partial derivatives. Which is it?

8. Oct 17, 2007

nrqed

Just to correct a typo...This should read

$$ln(\sqrt{x^2+y^2}) = ln((x^2+y^2)^\frac{1}{2}) = \frac{1}{2}ln(x^2+y^2)$$