Question

[CALCULUS III—DIFFERENTIATION] Given f(x, y) find the total differential of f(x, y). figure inside.

Answer 1

\[z=xsin(y/x)\]

Answer 2

http://tutorial.math.lamar.edu/Classes/CalcIII/Differentials.aspx

Answer 3

\[dz=f_xdx+f_ydy\]

Answer 4

\[u=x,\ du=dx\]\[v=sin(y/x),\ dv=-cos(y/x)/x^2\]

Answer 5

\[f_x=-xcos(y/x)/x^2+sin(y/x)\]\[f_x=-x^{-1}cos(y/x)+sin(y/x)\]

Answer 6

\[u=x,\ du=0dy\]\[v=sin(y/x),\ dv=x^{-1}cos(y/x)\]\[f_y=cos(y/x)\]

Answer 7

I think I may have made a mistake on fy, I am going to verify on paper.

Answer 8

\[f_x=-y/x\ cos(y/x)+sin(y/x)\]

Answer 9

\[dz=-y/x\ cos(y/x)+sin(y/x)+cos(y/x)\]

Answer 10

@calmat01 verification?

Answer 11

Apparently I am missing something because I should get a two term numerator over x

Answer 12

let me check. I will go through it here in a sec.

Answer 13

ok...checking now.

Answer 14

apparently my total differential should have "dx"

Answer 15

see attachment:

I double checked and I am leaning toward
\[dz=\frac{-y\ cos(y/x)+x\ sin(y/x)}{x}dx+\frac{x\ cos(y/x)}{x}dy\]

Answer 16

ok, that dx term should have an x^2 in the denominator because the derivative of (y/x) holding y constant is (-y/x^2)

Answer 17

Good eyes. I had it squared initially, but dropped it to a first degree in my work.

Answer 18

Wait, I reduced it because I had x / x^2

Answer 19

Everrything else looks good...checking with my work now.

Answer 20

hang on.

Answer 21

verification: http://www.wolframalpha.com/input/?i=partial+derivative+respect+to+x+x%28sin%28y%2Fx%29%29+

see attachment

Answer 22

I verified the dy as well. The answer I posted for dz with the attachment is correct.

Answer 23

yep, caught my mistake. It is correct...f_x that is..checking f_y now.

Answer 24

f_y is much easier.

Answer 25

yep, everything confirms with what I obtained.

Answer 26

Good, looks like you are getting the hang of it.

Answer 27

Checking on a struggler, brb.