# Thread: Finding the partial derivatives for ...

1. ## Finding the partial derivatives for ...

f(x,y)= arctan(y/x)

Im familiar with "outer" and "inner" derivates and usually know how to work them. I also know that the derivative of arctan = 1/(1+x^2)

so i get

f'x= 1/(1+x^2) * (y/x) * ??

What is the x-derivative of y/x ?

I also cannot seem to find the partial derivates to:

f(x,y)= (x+y)/(x-y)

Help would be highly appreciated...

2. When you calculate the partial derivative with respect to x, you treat y as a constant. Likewise when you differentiate with respect to y.

So you get:
$\displaystyle \displaystyle f(x,y) = arctan \left( \frac{y}{x} \right) \Rightarrow f_y(x,y) = \frac{1}{1+ \left( \frac{y}{x} \right) ^2} \cdot \overbrace{\frac{\partial}{\partial y} \left( \frac{y}{x} \right)}^{ = \frac{1}{x}}$ $\displaystyle \displaystyle = \frac{1}{x + \frac{y^2}{x}} \cdot \frac{x}{x} = \frac{x}{x^2 + y^2}$

You do the other one.

3. To find $\displaystyle \frac{\partial f}{\partial x}$, take the derivative keeping $\displaystyle y$ constant.

So let $\displaystyle u = \frac{y}{x}$ so that $\displaystyle f = \arctan{u}$.

$\displaystyle \frac{\partial u}{\partial x} = -\frac{y}{x^2}$.

$\displaystyle \frac{df}{du} = \frac{1}{1 + u^2}$

$\displaystyle = \frac{1}{1 + \left(\frac{y}{x}\right)^2}$

$\displaystyle = \frac{1}{1 + \frac{y^2}{x^2}}$

$\displaystyle = \frac{1}{\frac{x^2 + y^2}{x^2}}$

$\displaystyle = \frac{x^2}{x^2 + y^2}$.

Therefore $\displaystyle \frac{\partial f}{\partial x} = -\frac{y}{x^2}\left(\frac{x^2}{x^2 + y^2}\right)$

$\displaystyle = -\frac{y}{x^2 + y^2}$.

To find $\displaystyle \frac{\partial f}{\partial y}$, take the derivative with respect to $\displaystyle y$ keeping $\displaystyle x$ fixed.

Let $\displaystyle u = \frac{y}{x}$ so that $\displaystyle f = \arctan{u}$.

$\displaystyle \frac{\partial u}{\partial y} = \frac{1}{x}$.

$\displaystyle \frac{df}{du} = \frac{1}{1 + u^2}$

$\displaystyle = \frac{1}{1 + \left(\frac{y}{x}\right)^2}$

$\displaystyle = \frac{x^2}{x^2 + y^2}$.

Therefore $\displaystyle \frac{\partial f}{\partial y} = \frac{1}{x}\left( \frac{x^2}{x^2 + y^2}\right)$

$\displaystyle = \frac{x}{x^2 + y^2}$.

4. Originally Posted by tinyone
f(x,y)= arctan(y/x)

Im familiar with "outer" and "inner" derivates and usually know how to work them. I also know that the derivative of arctan = 1/(1+x^2)

so i get

f'x= 1/(1+x^2) * (y/x) * ??

What is the x-derivative of y/x ?

I also cannot seem to find the partial derivates to:

f(x,y)= (x+y)/(x-y)

Help would be highly appreciated...
$\displaystyle f(x, y) = \frac{x + y}{x - y}$.

$\displaystyle \frac{\partial f}{\partial x} = \frac{(x - y)\frac{\partial}{\partial x}(x + y) - (x + y)\frac{\partial}{\partial x}(x - y)}{(x - y)^2}$

$\displaystyle = \frac{(x - y) - (x + y)}{(x - y)^2}$

$\displaystyle = -\frac{2y}{(x - y)^2}$.

$\displaystyle \frac{\partial f}{\partial x} = \frac{(x - y)\frac{\partial}{\partial y}(x + y) - (x + y)\frac{\partial}{\partial y}(x - y)}{(x - y)^2}$

$\displaystyle = \frac{(x - y) + (x + y)}{(x - y)^2}$

$\displaystyle = \frac{2x}{(x - y)^2}$.

5. Thank you both so much, this forum is amazing.

PS. If anybody else is eager to help Ive got another problem I can't solve

Determine x*(f'x)+y*(f'y)+z*(f'z)

if : f(x,y,z)= ln(x^3+y^3+^z^3 - 3xyz)

The answer is simply 3. But when I derivate with regards to x, y and z and put it into x*(f'x)+y*(f'y)+z*(f'z) I get something far more complicated which I cannot simplify.