What are the partial derivatives of the given functions with respect to x and y?

[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 don't 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 don't even know how to do this one.

 

[cookiemonster]

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.

cookiemonster

 

[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.

 

[cookiemonster]

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.

cookiemonster

 

[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.

 

[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
$\frac{\partial f}{\partial y}= \frac{-x}{x^2+ y^2}$