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

- anonymous

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

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- chestercat

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- anonymous

Can 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

\[f(x,y)= \int\limits_{y}^{x} \cos (t^9) dt\]

- anonymous

this is what i got

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## More answers

- anonymous

\[1/x^9 \sin(x^9)+1/y^9 \sin(y^9) +C\]

- anonymous

not sure if its right though

- anonymous

ohhh..

- anonymous

i am suppose to integrate by parts right?

- anonymous

1/9sinx^9-1/9sinx^9+c

- anonymous

In 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

i totally spaced out on that

- anonymous

1/9sinx^9-1/9siny^9+c

- anonymous

Ok lets see, I believe your integral is right, although there's no need for the +c because you have definite bounds...

- anonymous

yes you are right no need +c

- anonymous

so we don't integrate this by u substitution?

- anonymous

or some other integration method?

- anonymous

no need substitution

- anonymous

hmm.......... actually

- anonymous

Definite 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

So 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

so what does that mean?

- anonymous

You 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

http://www.wolframalpha.com/input/?i=integral%28cos%28t^9%29dt%29

- anonymous

Hmmm 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

yeah we haven't learned this and there isn't an example in a book

- anonymous

http://how-to-spell-ridiculous.com/

- anonymous

You are going to need an incomplete gamma or knowledge of the Heaviside to solve this.

- anonymous

hahahaha thanks ScienceWiz

- anonymous

<3

- anonymous

So 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

So 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

yes

- anonymous

where F(x)= integral of x

- anonymous

then 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

the 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

Thoughts?

- anonymous

yeah I am kind of confused because I am a bit rusty on the FTC

- anonymous

I'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

FTC just states The derivative of the Integral is your original equation..

- anonymous

Ya but you can't apply that theorem to this situation unless you were to integrate again, which would cause the same problem.

- anonymous

hmmm? 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

I need to find the first partial derivatives fx and fy

- anonymous

hmmm.. 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)=âˆ«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

so f(x) = cos (x ^9) and -f(y)= -cos(y^9)? because of FTC

- anonymous

Pretty much.

- anonymous

that seems much easier :) I appreciate your help!!
THANKS

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