# Partial derivative

1. May 26, 2004

### galipop

Hi All,

Can someone refresh my memory and show me how to find the following partial derivate:

$$t=\frac{x}{\sqrt{x^2+y^2}}$$

with respect to x.

Thanks

Last edited: May 26, 2004
2. May 26, 2004

### arildno

The partial derivative of t(x,y) with respect to x is found by differentiating the expression as in 1-variable calculus; just remember that y is to be considered a constant. For example:
$$t(x,y)=xy\rightarrow\frac{\partial{t}}{\partial{x}}=y$$

3. May 26, 2004

### galipop

Yeah I understand the concept. It's just that I've forgotten how to find the derivative of a quotient. The textbook that I have doesn't explain it clearly.

4. May 26, 2004

### arildno

The simplest way is to solve for the derivative of t w.r. to x by differentiating the equivalent equation with respect to x:
$$\sqrt{x^{2}+y^{2}}t(x,y)=x$$

Another easy way is (I'll give it for one variable):
$$t(x)=\frac{f(x)}{g(x)}$$
Use the product rule and the chain rule:
$$\frac{dt}{dx}=\frac{f'(x)}{g(x)}-\frac{f(x)}{g^{2}(x)}*g'(x)$$

Last edited: May 26, 2004
5. May 26, 2004

### TALewis

I remember the quotient rule as "Lo d Hi minus Hi d Lo over Lo squared," or:

$$\frac{d}{dx}\,\frac{f(x)}{g(x)}=\frac{gf' - fg'}{g^2}$$

Last edited: May 26, 2004
6. May 26, 2004

### NateTG

Personally I generally use:
$$\frac{a}{b}=a (b)^{-1}$$
so in your case
$$\frac{x}{\sqrt{x^2+y^2}}=x (x^2+y^2)^{-\frac{1}{2}}$$

7. May 26, 2004

### e(ho0n3

When I forget how do calculate a particular derivate I always derive it by using the definition of the derivate, i.e.
$$\frac{d}{dx}f(x) = \lim_{h \to \infty} \frac{f(x+h)-f(x)}{h}$$​
e(ho0n3

8. May 26, 2004

### galipop

How do you proceed using this form?

9. May 26, 2004

Product rule and power rule.

f'*g + g'*f, where f = x and g = (x^2 + y^2)^(-1/2), so f' = 1 and g' = -x(x^2 + y^2)^(-3/2)

And e(ho0n3, don't you mean limit as h goes to 0?