Derivative of a function with ln

[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!

 

[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?

 

[ns5032]

Yes I believe so...

Would it then be...

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

...if I remember correctly??

 

[ns5032]

Which would then simplify to

1/(x+y)

?

 

[supraanimo]

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

 

[rock.freak667]

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

 

[HallsofIvy]

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?

 

[nrqed]

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

Is it easier now?

Just to correct a typo...This should read

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