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anonymous
 5 years ago
HELP:find the first partial derivative of f(x,y) integral from x to y cos(t^9) dt
anonymous
 5 years ago
HELP:find the first partial derivative of f(x,y) integral from x to y cos(t^9) dt

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you restate the question? You want to find the first order partial derivative of a function f(x,y) with respects to both x and y correct? What exactly is your function?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f(x,y)= \int\limits_{y}^{x} \cos (t^9) dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1/x^9 \sin(x^9)+1/y^9 \sin(y^9) +C\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not sure if its right though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am suppose to integrate by parts right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/9sinx^91/9sinx^9+c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In said situation, given theorized, ideal Pythagorean standards, one could, when, concluded extremities parametize area, fortitude of x with respect to inifinite limit gives, under mathematical conditions, that, stated integral has an unreal answer if eliminated factors cancel in a forthright fashion. All is explained by science

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i totally spaced out on that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/9sinx^91/9siny^9+c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok lets see, I believe your integral is right, although there's no need for the +c because you have definite bounds...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes you are right no need +c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we don't integrate this by u substitution?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or some other integration method?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm.......... actually

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Definite C, when referring to tangency, is, without question, derivized by cascadation of three bounds in the three dimensional coordinate plane, as defined by Cartesian math as well as Mendel, as, with respect to z, a common factor is pulled, given that a trifactor of polynomic trigonometry is frozen in intervals of, cosine, when not shown to be aligned, yields a C of zero which can, ultimately, not need be shown unless graphing is necessary.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So we got 1/x^9 sin(x9)+1/y^9 sin(y9) when evaluated between x and y.... This means our integral without evaluating it, and in terms of t (aka indefinite integral) was.... 1/t^9 cos(t^9)... if you derive that in terms of t you do not get the original equation...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what does that mean?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You are either missing part of the problem or your teacher is an retriceole. You can't solve this easily, if at all, because you are integrating for cosine with a function inside and nothing outside to return upon integrating. Try using a trig identity if you can think of one, or www.wolframalpha.com/. Otherwise you can just say unable to solve because it seems unreasonable for you to be asked to solve that if your teacher hasn't shown you how.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=integral%28cos%28t^9%29dt%29

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmmm i actually had a problem just like this on my vector calc test a few weeks ago... I didn't get the problem but i think it has to do with cos(x^2) being an odd function... I'm looking for some sort of trig identity or trick now because neither u sub or by parts will work for something like this, and wolfram gives me a rediculious answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah we haven't learned this and there isn't an example in a book

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You are going to need an incomplete gamma or knowledge of the Heaviside to solve this.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hahahaha thanks ScienceWiz

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So I think I know the answer... If we use the Fundamental Theorem of Calculus we can state that the Integral of your original function then derivative in respects to t is just the original function right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So if you take the Integral of your original function f(x) then evaluate with respects to x and y. You'll get some F(x)F(y) right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where F(x)= integral of x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then we take the partial with respects to x of F(x)F(y). F(y) goes to 0 because there are no x terms, while F(x) goes to the original function just in terms of x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the same applies for your partial derivative of y... So you should just get for your\[f _{x}=\cos(x^9)\] and\[f _{y}=\cos(y^9)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah I am kind of confused because I am a bit rusty on the FTC

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm having a hard time following exactly what you are saying. The main issue that I see is there is no way to get around integrating for t at somepoint. And you can't do that at the level of knowledge you are at right now. Also this link says no. And if you delete "from x to y" you can look at the indefinite integral. http://www.wolframalpha.com/input/?i=integral%28cos%28t^9%29dt%29+from+x+to+y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0FTC just states The derivative of the Integral is your original equation..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ya but you can't apply that theorem to this situation unless you were to integrate again, which would cause the same problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm? the initial problem states "find the first partial derivative of f(x,y) integral from x to y cos(t^9) dt" doesn't that mean you want to find the partial derivatives? Or do you just want the integral?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I need to find the first partial derivatives fx and fy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm.. ok let me try explaining my logic better... So FTC states...\[d/dx(\int\limits_{a}^{x}f(t)dt) :=F'(t)= f(x)\] this means after you take the integral if you take the derivative you get your original function... So our function is f(x,y)=∫yxcos(t9)dt\[f(x,y)=\int\limits_{y}^{x}\cos(t^9) dt= \int\limits_{y}^{x}f (t)dt\] We can just rewrite our\[ \int\limits_{y}^{x}f (t)dt= F(x)F(y) \] where F(x) is equal to the integral of our original equation evaluated at x... So now we take out function F(x)F(y) and take the partials in respects to x and y...\[f _{x}(F(x)F(y))= f(x)\] and \[f _{y}(F(x)F(y))= f(y)\] This is because when we take the partial of F(y) with respects to x it goes to 0...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so f(x) = cos (x ^9) and f(y)= cos(y^9)? because of FTC

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that seems much easier :) I appreciate your help!! THANKS
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