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In this question, I'm asked to find the differentials of each term, how do I do this? \[y^3 - xt - 1 = x^2 + y^2\]
help still tengo
1sec i think i did this before
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,
There's a function here?
" you are also Given a function"
ok what u think function is
Something that isn't in this equation....?
the medal ?
is it fixed medal accdeint
I don't know what you're talking about.
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 ?
\[y^3 - xt - 1 = x^2 + y^2\] take differential through out \[3y^2dy - (dx*t+x*dt) = 2xdx+2ydy\] grouping terms we get \[dy(3y^2-2y) + (-t-2x)dx -xdt=0\]
\[y^3 - xt - 1 = x^2 + y^2\] take differential through out \[3y^2dy - (dx*t+x*dt) = 2xdx+2ydy\] grouping terms we get \[(3y^2-2y)dy + (-t-2x)dx -xdt=0\]
makes sense i get it
So about the term \[y^3\], why does it become \[3y^2 dy\] ?
whats the derivative of \(u^3\) with respect to \(u\) ?
3u du? "with respect to u," what does it mean?
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/
Oh wait. Lemme think. So if I take the differentials of this equation: \[y^2 - xy = \sin x\] I'll end up with \[2 ydy - xdy - ydx = cosx dx\]
Yes that looks good! but still could you please watch that video, that video explains differentials a bit more accurately...
Whew. I'm terrible with differentials. x_x Thanks! I will!
I'm sure that video clears up everything! yw :)