| Joel
Kammet |
I am to find the first partial derivatives of f(x,y)=òxy cos(t2) dt are they simply fx=-2x cos(x2) and fy=2y cos(y2) ? If not, please explain how to do this. |
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| Joel
Kammet |
That was supposed to be: f(x,y) = cos(t2
)dtI'm pretty confused by this. Isn't this a composite function that calls for use of the chain rule? I'm thinking that I have g(t) = t2 and the integral is f(g(t)) which gets evaluated at t=y and t=x, thus becoming f(g(x)) and f(g(y)), so cos(y2 ) - cos(x2 ) is df/dg and I still have to differentiate that wrt x to get f/ x and
wrt y to get f/ y.So, f/ x =
df/dg* g/ xand f/ y =
df/dg * g/ xThen, f/ x =
/ x[cos(y2
) - cos(x2 )] = -2xcos(x2 )and f/ y =
/ y[cos(y2
) - cos(x2 )] = 2ycos(y2 )But I'm not very confident about this. Please tell me if it's all nonsense. |
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| Andre
Rzym |
It's not correct, I'm afraid. Try writing ¶f/¶x (partial derivative) as a limit as D® 0, and substitute the integral into the limit expression. You should see that 'most of' the integral cancels (you have the difference of two integrals which just differ by their limits), and you are then interested in evaluating an integral from x to x+D. I'll post more later if that makes no sense. Andre |
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| Kerwin
Hui |
There should not be a factor of 2x (or 2y respectively). If you write the integral in the form ![]() for some fixed point p, the question reduces to applying the fundamental theorem of calculus, i.e. ![]() and similar. Kerwin |
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| Joel
Kammet |
OK. I really needed to go back & think about the fundamental theorem for a while. The "difference" aspect is clear. I was just getting confused about which function I was dealing with. i.e. ![]() ![]() vs. ![]() ![]() and ending up with a mishmash of the two. I've had better moments. Thanks, both of you, for your help. |