Recent content by Joeda

ok once again thanks Voko you have been brilliant! f(x)= [1+(x^2+y^2)^{-1/2}(x)] [\frac{1}{x+\sqrt{x^2+y^2}} - \frac{1}{x-\sqrt{x^2+y^2}}] So this is written correctly then?

As I had f(x)= x+(x^2+y^2)^{1/2} f(x)= 1+\frac{1}{2}(x^2+y^2)^{-1/2}(2x) f(x)= 1+(x^2+y^2)^{-1/2}(x) and f(x)= x-(x^2+y^2)^{1/2} f(x)= 1-\frac{1}{2}(x^2+y^2)^{-1/2}(2x) f(x)= 1-(x^2+y^2)^{-1/2}(-x) As -\frac{1}{2} * 2x = -x

So we get f(x)= [1+(x^2+y^2)^{-1/2}(x)] [\frac{1}{x+\sqrt{x^2+y^2}} - \frac{1}{x-\sqrt{x^2+y^2}}] That doesn't look right as we had (x) and (-x) in the above equation.

So I get f(x)=\frac{1}{x+\sqrt{x^2+y^2}} . 1+(x^2+y^2)^{-1/2}(x) - \frac{1}{x-\sqrt{x^2+y^2}} . 1-(x^2+y^2)^{-1/2}(-x)

So can this expression be simplified further by multiplying the x term outside the brackets by the terms inside the brackets, or should I leave these terms as is and just plug in my values for x & y? I would think you can't but just wanted to clarify.

Wow I can't believe I didnt see that! Amazing how you can get caught up in the more difficult aspects of a question and miss the basics! Thankyou Voko for your patience its very much appreciated.

Sorry Sir once again I am lost :) If I have (x+y^2)^3 using the chain rule I get \frac{df}{dx}=3(x+y^2)^{2} as derivative of x = 1 so this could be written as \frac{df}{dx}=3(x+y^2)^{2} *(1) and \frac{df}{dy}=3(x+y^2)^{2}*(2y) \frac{df}{dy}=6y(x+y^2)^{2} So then shouldnt...

Well the 3 came from the derivative of x with respect to x = 1 1+1\2=3/2 The derivative of x with respect to y is zero?

Derivative of x with respect to x is 1. So its 1+\frac{1}{2}(x^2+y^2)^{-1/2}(2x) \frac{3}{2}(x^2+y^2)^{-1/2}(2x) Please tell me this is correct!

So does x+(x^2+y^2)^{1/2} equal x+\frac{1}{2}(x^2+y^2)^{-1/2}(2x) ?

voko I am lost now.

Is the power outside of the brackets not required to be derived? Also if x & y are inside brackets when taking the partial derivative of x, isn't the derivative of x taken and y is just a constant? ie. f(x) = {(x^2+y^2)} = 2x + y^2 whereas if it was f(x) = {x^2+y^2} = 2x

Thanks everyone for all your help. So... f(x,y)=ln|\frac{x+\sqrt{x^2+y^2}}{x-\sqrt{x^2+y^2}}| f(x,y)=ln|(x+\sqrt{x^2+y^2})| - ln|(x-\sqrt{x^2+y^2})| f(x)=\frac{1}{x+\sqrt{x^2+y^2}} . \frac{3}{2}(2x+y^2)^{-1/2} - \frac{1}{x-\sqrt{x^2+y^2}} . \frac{1}{2}(2x+y^2)^{-1/2}

ok I am really starting to get confused here. How about this? f(x,y)=In|x+√(x^2+y^2)|-In|x-√(x^2+y^2)| f(x,y)=(1/2)In|x+(x^2+y^2)|-(1/2)In|x-(x^2+y^2)| f(x)=(1/x)(1/2)(1/(x+x^2+y^2))(1+2x+y^2)-(1/x)(1/2)(1/(x-x^2+y^2))(1-2x+y^2)...

ok hows this look? Dividing both numerator and denominator by x f(x,y)= In|(√(x^2+y^2))/(√(x^2+y^2))| Simplifying using Log Rules f(x,y)= In|(√(x^2+y^2))|-In|(√(x^2+y^2))| f(x,y)= (1/2)In|((x^2+y^2))|-(1/2)In|((x^2+y^2))| f(x) = (1/2)(1/(x^2+y^2)-(1/2)(1/(x^2+y^2)...