Multivariable Calculus: Partial Differentiation: f(x,y,z) = (x^3)(y^2)(z^4) + 2xy + z?

The same thing happened to all three of them - they were eliminated because they are constants. You will recall from ordinary differentiation that constants simply disappear when differentiated (i.e. if f(x)=x²+2x+1, f'(x)=2x+2 - the derivative of 1 is 0 and is therefore absorbed). In partial derivatives, every variable except the one you are differentiating with respect to is assumed to be constant and treated as such. Thus, d(z)/dx simply reduces to 0. Similarly when differentiating with respect to y. And In the final problem, 2xy is the product of three constants and is therefore a constant.

Edit: Actually, this reminds me of an old joke I heard while I was in high school. e^x was walking down the street when suddenly he heard a commotion. When he looked around to see what it was, 3 slammed into him at full speed, nearly knocking him over. He asked 3 what he was running away from, and 3 said "Didn't you hear? There's a differential operator coming here. He'll differentiate me and make me disappear. You should run, before he gets you too." e^x laughed at this, saying "Ha! Why should I be afraid, I'm e^x! I won't be changed one bit!." So e^x boldy walked down the street in the opposite direction that everyone else was running, until at last he saw the guy that everyone was running away from. And he said to him: "So, you're that differential operator everyone's so scared of. Well, you can try your worst on me, I am not afraid." And the differential operator said to him "well, you should be. I'm d/dy"

ok. so for 1, when you differentiate with respect to x, you let y and z be constants. call them c and d if you want.

so then the equation would be

(x^3)(c^2)(d^4) + 2xc + d

and you can write (c^2)(d^4) as another constant. constant as in a number. you can call it A. so then the equation is just

Ax^3 + 2cx + d. so differentiate the first term as you would if it was like 5x^3. or something

d/dx (5x^3) = 3*5 x^2

so d/dx (Ax^3) = 3*A x^2

and then teh 2nd term is 2c and d is just a constant. so its like taking derivative of 2. u get 0. thats why the last z term disappears. so the derivative is

3Ax^2 + 2c. and now you gotta put all the constants back in terms of y and z. and you get the answer for 1.

2. same thing here, but you gotta put constants for the x and z this time and the last one is same with x and y constants.

let me know fi u still need more help

for #1, since you are taking the derivative with respect to 'x', 'z' can be considered a constant, thus the derivative of a constant = 0, which causes the 'z' term to not be in the answer.

The same reasoning can be deduced for #2 and #3. In #2, 'z' can be treated as a constant since you are taking the derivative with respect to y. In #3, the derivative of 2xy = 0, because you are taking the derivative with respect to 'z'.

A general rule with partial differentiation is to treat the variables as constants if you are not taking the partial derivative of that variable.

Hope this helps

#1: If you differentiate f with respect to x, z is a constant. All constants disappear when you differentiate

#2: See #1

#3: See #1

If this is hard to understand, try to call the constants c1 and c2 instead. That usually helps, e.g.

(x^3)(c1^2)(c2^4)+2c1x + c2 when you diff. with respect to x