jojokiw3
  • jojokiw3
Help with differentials.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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jojokiw3
  • jojokiw3
In this question, I'm asked to find the differentials of each term, how do I do this? \[y^3 - xt - 1 = x^2 + y^2\]
jojokiw3
  • jojokiw3
@Directrix
anonymous
  • anonymous
help still tengo

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anonymous
  • anonymous
EspaƱol
jojokiw3
  • jojokiw3
@zepdrix
anonymous
  • anonymous
1sec i think i did this before
anonymous
  • anonymous
if you Note that if we are just given then the differentials are df and dx and we compute them in the same manner. you are also Given a function we call dy and dx differentials and the relationship between them is given by,
jojokiw3
  • jojokiw3
There's a function here?
anonymous
  • anonymous
what
jojokiw3
  • jojokiw3
" you are also Given a function"
anonymous
  • anonymous
ok what u think function is
jojokiw3
  • jojokiw3
Something that isn't in this equation....?
anonymous
  • anonymous
the medal ?
anonymous
  • anonymous
is it fixed medal accdeint
jojokiw3
  • jojokiw3
I don't know what you're talking about.
anonymous
  • anonymous
im trying to guess what function that has nothing to do with equation u are talking about so i thought was the medal so can u EXPLAIN WHAT UR TALKING ABOUT ?
ganeshie8
  • ganeshie8
\[y^3 - xt - 1 = x^2 + y^2\] take differential through out \[3y^2dy - (dx*t+x*dt) = 2xdx+2ydy\] grouping terms we get \[dy(3y^2-2y) + (-t-2x)dx -xdt=0\]
ganeshie8
  • ganeshie8
\[y^3 - xt - 1 = x^2 + y^2\] take differential through out \[3y^2dy - (dx*t+x*dt) = 2xdx+2ydy\] grouping terms we get \[(3y^2-2y)dy + (-t-2x)dx -xdt=0\]
anonymous
  • anonymous
makes sense i get it
jojokiw3
  • jojokiw3
So about the term \[y^3\], why does it become \[3y^2 dy\] ?
ganeshie8
  • ganeshie8
whats the derivative of \(u^3\) with respect to \(u\) ?
jojokiw3
  • jojokiw3
3u du? "with respect to u," what does it mean?
ganeshie8
  • ganeshie8
I suggest you watch this video http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-b-chain-rule-gradient-and-directional-derivatives/session-32-total-differentials-and-the-chain-rule/
jojokiw3
  • jojokiw3
Oh wait. Lemme think. So if I take the differentials of this equation: \[y^2 - xy = \sin x\] I'll end up with \[2 ydy - xdy - ydx = cosx dx\]
ganeshie8
  • ganeshie8
Yes that looks good! but still could you please watch that video, that video explains differentials a bit more accurately...
jojokiw3
  • jojokiw3
Whew. I'm terrible with differentials. x_x Thanks! I will!
ganeshie8
  • ganeshie8
I'm sure that video clears up everything! yw :)

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