# Partial derivatives

1. Mar 29, 2004

### phy

ok i need help with a few questions. i'll post the question first and then what i get as an answer, the first one is the partial derivative with respect to x and the second one with respect to y. these are the even number questions from my textbook and they dont have answers to them so if someone could check them, i would really appreciate it; thanks.

a) z = yln|x|
b) z = x^y
c) z = xy^2/(x^2+y^2)
d) z = arctan(x/y)

a) 1/x, 1
b) yx^(y-1), (x^y)(ln|x|)
c) [(x^2)(x^2+y^2)-xy^2(2s)]/(x^2+y^2)^2,
[(2xy)(x^2+y^2)-st^2(2s)]/(x^2+y^2)^2
d) i dont even know how to do this one.

2. Mar 29, 2004

Be sure to treat the part of the expression that doesn't have the variable you're differentiating with respect to as a constant.

For the arctan problems, take the tangent of both sides and differentiate implicitly.

3. Mar 29, 2004

### phy

um, do you think you could explain that? thanks. i've read over my profs notes and they make sense but they are the simple examples.

4. Mar 29, 2004

Okay. In the first problem,

z = y ln|x|

If we're differentiating with respect to x, we're going to treat anything with a y in it as constant. That is, we're going to treat y multiplier in front as a constant. So essentially we're differentiating

z = c*ln|x|

where c is some constant. I'm sure you know how to differentiate that to get

dz/dx = c/x = y/x (since c = y).

Now to differentiate with respect to y, we treat anything with an x as a constant. That would be the expression ln|x|. So we're differentiating

z = y*c

which is

dz/dy = c = ln|x| (since c = ln|x|)

For the last one, you need to take the tangent of each side to get the equation

tan(z) = x/y

We then use the chain rule to differentiate tan(z), i.e.

$$\frac{\partial f(z)}{\partial x} = \frac{\partial f(z)}{\partial z}\frac{\partial z}{\partial x}$$

it is the same for differentiating with respect to y, just replace the x's in the above equation with y's.

That gives us the left-hand side. The right-hand side you differentiate normally. Once you have the left- and right-hand sides, solve for dz/dx (or dz/dy, if you're differentiating with respect to y) and that will be your answer. If you have an expression with z in your answer, substitute it away using the problem statement.

5. Mar 30, 2004

### philosophking

Hm...

If we have d/dx [tan x]...doesn't that equal 1/(1+x^2) ??? you should know that from calc1 or 2 right? So just apply the same form to tan x/y, holding y constant for (partial x) and x costant for (partial y). I think this would be right:

(partial x) = (1/y)/(1+(x/y)^2)

and

(partial y) = (-x/y^2)/(1+(x/y)^2)

Check my work though, not totally sure.

6. Mar 30, 2004

### HallsofIvy

"(partial x) = (1/y)/(1+(x/y)^2)"

Yes, although it might be a good idea to multiply both numerator and denominator by y2 to get $\frac{\partial f}{\partial x}= \frac{y}{x^2+y^2}$

"(partial y) = (-x/y^2)/(1+(x/y)^2)"

Excellent! Now multiply both numerator and denominator by y2 to get
$\frac{\partial f}{\partial y}= \frac{-x}{x^2+ y^2}$

Last edited by a moderator: Mar 30, 2004