Finding the partial derivatives for ...

[OrangeOne]

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...

 

[galactus]

The first one, y is a constant because we are differentiating w.r.t x.

Say $\displaystyle z=tan^{-1}(\frac{y}{x})$

Let $\displaystyle u=\frac{y}{x}$

$\displaystyle tan^{-1}(u)$

Then, $\displaystyle \frac{dz}{dx}=\frac{dz}{du}\cdot \frac{du}{dx}$

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

$\displaystyle \frac{du}{dx}=\frac{-y}{x^{2}}$

So, $\displaystyle \frac{dz}{du}\cdot \frac{du}{dx}=\frac{1}{u^{2}+1}\cdot \frac{-y}{x^{2}}=\frac{-y}{(u^{2}+1)x^{2}}$

Now, sub $\displaystyle u=\frac{y}{x}$ back in:

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

See there?. Does that help?.

 

[OrangeOne]

Thanks, that cleared up a lot.

I also found the solution to
f(x,y)= (x+y)/(x-y)

However, Im now working with the chain rule and already having problems solving this one:

f(x,y)= e^((x^2)*y)) + xy
u(t)= f(cos t,sin t)

Determine u(t) and u'(t)