Question

find \[\frac{ \delta z }{ \delta x }\]and\[\frac{ \delta z }{ \delta y }\]where\[\frac{ yz }{ x }-\sin x \cos^{2} y =ye^{yz}-z(xy+1)^{3} \]\[z = f(x,y)\]

Answer 1

Can you take the partial derivative of the first term with respect to x?

Answer 2

erm, do u mean del/del x (yz/x)?

Answer 3

yesh

Answer 4

ya, no prob wait, i write the eqn

Answer 5

del/del x (yzx^(-1))=(del z/del x )yx^(-1) + (-1)yzx^(-2)

Answer 6

= (del z / del x)(y/x) -yz/(x^(2)) right?

Answer 7

yeah that is correct, good work ,

now , the other terms

Answer 8

(del / del x )(sin (x) cos ^2(y)) = cos (x) cos^2 (y)

Answer 9

what does Del mean here

Answer 10

, delta ?

Answer 11

∂, partial derivative \(\partial \) \partial

Answer 12

ok thanks

Answer 13

once you have take the implicit partial derivative of the both sides of the equation, you just have to rearrange to solve for \(\dfrac{\partial z}{\partial x }\)

Answer 14

(del / del x)ye^(yz)= (ye^(yz))((del z/ del x)(y))=(del z/ del x)y^2e^(yz)

Answer 15

i still hv some problem, that is my ans and ans in wolframalpha is different, is that my ans wrong?? i'll show all steps... but i cant find any exponent in wolframalpha

Answer 16

maybe wolfram doesn't know that z(x,y)

Answer 17

i dont think my solution can be simplify some more... but wolfram seem like have some way to simplify

Answer 18

the original equation is
y*x^(-1)*z - sin x * cos^2(y) = y*e^(y*z)- z(xy+1)^3

Answer 19

ya

Answer 20

partial z with respect to x (implicitly is)
I will use z'_x for partial z with respect to x
y(-1)x^(-2)z + y(x^(-1)z'_x = y*e^(y*z)*y*z'_x - z'_x(xy + 1)^3 -z(3)(xy+1)^2*y

Answer 21

@neoh147 , i got the same as you on that .png

\[\frac{\partial z}{\partial x}=z_x=\dfrac{\frac{yz}{x^2}+\cos x\cos^2y-3zy(xy+1)^2}{\frac yx-y^2e^{yz}+(xy+1)^3}\]

Answer 22

I can do it in maple, one moment

Answer 23

ok, that means probably my calculation is nothing wrong... can u help me to check the partial diff respect to y??

Answer 24

I can give you maple solution, how do upload a .png?

Answer 25

use print screen, paste it on paint and attach file

Answer 26

one sec

Answer 27

ok ^_^

Answer 28

its ugly though ;)

Answer 29

ok, but my answer is require in the simplest form, not expanded form, anyway thx for sharing the eqn ^_^

Answer 30

@neoh147 the third line on your ∂/∂y

have yo got the right sign on your derivative of that trig function?

Answer 31

ya, suppose differentiate cos y should be - sin y, I make a careless mistake ><

Answer 32

and i think you missed a \(y\) also


\[\frac{\partial z}{\partial y}(yx^{-1}-y^{\color{red}2}e^{yz}+...\]

Answer 33

ok >w<

Answer 34

going back to the first one ,
if you want to simplify\[\frac{\partial z}{\partial x}=\dfrac{\frac{yz}{x^2}+\cos x\cos^2y-3zy(xy+1)^2}{\frac yx-y^2e^{yz}+(xy+1)^3}\]
multiply numerator and denominator by x^2
\[=\dfrac{yz+x^2\cos x\cos^2y-3x^2yz(xy+1)^2}{xy-x^2y^2e^{yz}+x^2 (xy+1)^3}\]

Answer 35

@perl can you ask maple to factorise your results,?

Answer 36

let me try
just type factor?

Answer 37

wont let me

Answer 38

ok, i get it d, the complex fractions must be reduce to a single fraction right?

Answer 39

i don't know maple @perl ,

but i think that you can factor you results they will look much more similar to the results that have been derived manually

Answer 40

i dont know how to simplify any further

Answer 41

ok thx a lot @UnkleRhaukus and @perl ^_^

Answer 42

partial z / partial x looks right

Answer 43

maple just factored out one x from the denominator, i guess the only common term