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Partial differentiation q

How do u do q8? Im guessing you dont equate w to u because it wouldn't really make sense. Here is what I have.

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Reply 2
Avatar for ztibor
ztibor
10
Original post by cooldudeman
How do u do q8? Im guessing you dont equate w to u because it wouldn't really make sense. Here is what I have.

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Inr q8 a function with 2 variables is given for you
Write down the first and second order partial derivatives of u according to
the question. (q8 is independent from eg q7)
Original post by ztibor
Inr q8 a function with 2 variables is given for you
Write down the first and second order partial derivatives of u according to
the question. (q8 is independent from eg q7)


This?

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Reply 5
Avatar for ztibor
ztibor
10
Original post by cooldudeman



∂2f∂x∂y+f=∂∂x(dfdu⋅∂u∂y)+f=\displaystyle \frac{\partial ^2 f}{\partial x\partial y} +f=\frac{\partial}{\partial x}\left (\frac{df}{du}\cdot \frac{\partial u}{\partial y}\right )+f=

Now use the product rule to diffeentiate wrt. x


=∂2f∂u2⋅∂u∂x⋅∂u∂y+dfdu⋅∂2u∂y∂x+f==\displaystyle \frac{\partial ^2 f}{\partial u^2} \cdot \frac{\partial u}{\partial x} \cdot \frac{\partial u}{\partial y}+\frac{df}{du} \cdot \frac{\partial ^2 u}{\partial y \partial x}+f=

.......
Reply 6
Avatar for TeeEm
TeeEm
19
Original post by cooldudeman



Original post by ztibor
∂2f∂x∂y+f=∂∂x(dfdu⋅∂u∂y)+f=\displaystyle \frac{\partial ^2 f}{\partial x\partial y} +f=\frac{\partial}{\partial x}\left (\frac{df}{du}\cdot \frac{\partial u}{\partial y}\right )+f=

Now use the product rule to diffeentiate wrt. x


=∂2f∂u2⋅∂u∂x⋅∂u∂y+dfdu⋅∂2u∂y∂x+f==\displaystyle \frac{\partial ^2 f}{\partial u^2} \cdot \frac{\partial u}{\partial x} \cdot \frac{\partial u}{\partial y}+\frac{df}{du} \cdot \frac{\partial ^2 u}{\partial y \partial x}+f=

.......



I got u times the (second derivative of f w.r.t u) + f =0

[apologies but I passionately refuse to use latex]
Original post by TeeEm
I got u times the (second derivative of f w.r.t u) + f =0

[apologies but I passionately refuse to use latex]


Isn't that what I got? Above post?
Reply 8
Avatar for TeeEm
TeeEm
19
Original post by cooldudeman
Isn't that what I got? Above post?


sorry I did not actually look at neither of your solutions

I did mine very quickly

Thanks for the PDF correction

( I tippex it out and when it dried i wrote again the original wrong answer. That is why it was correct later on)
Original post by ztibor
∂2f∂x∂y+f=∂∂x(dfdu⋅∂u∂y)+f=\displaystyle \frac{\partial ^2 f}{\partial x\partial y} +f=\frac{\partial}{\partial x}\left (\frac{df}{du}\cdot \frac{\partial u}{\partial y}\right )+f=

Now use the product rule to diffeentiate wrt. x


=∂2f∂u2⋅∂u∂x⋅∂u∂y+dfdu⋅∂2u∂y∂x+f==\displaystyle \frac{\partial ^2 f}{\partial u^2} \cdot \frac{\partial u}{\partial x} \cdot \frac{\partial u}{\partial y}+\frac{df}{du} \cdot \frac{\partial ^2 u}{\partial y \partial x}+f=

.......


I don't understand why u differentiated it again. What is wrong witg what I done?

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Reply 10
Avatar for ztibor
ztibor
10
Original post by cooldudeman
I don't understand why u differentiated it again. What is wrong witg what I done?

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this same what you got but your way was not correct . if I did not misread your solution
Reply 11
Avatar for TeeEm
TeeEm
19
Original post by cooldudeman
I don't understand why u differentiated it again. What is wrong witg what I done?

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Just send it.
hope it helps
Original post by TeeEm
Just send it.
hope it helps


Original post by ztibor
this same what you got but your way was not correct . if I did not misread your solution


This is what I got. Please can someone tell me if its fully correct.
(the pic on next post)
(edited 9 years ago)
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1414084944736.jpg46.8KB
Reply 14
Avatar for ztibor
ztibor
10
Original post by cooldudeman
This is what I got. Please can someone tell me if its fully correct.
(the pic on next post)


That is fully correct.

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