Question

find partial derivatives of F(x,y)= integral from x to y of (cos (e^t) dt.

Answer 1

take partial derivatives of \[\int\limits_{x}^{y}\cos (e ^{t})dt\]

Answer 2

ok let's solve ur problem

Answer 3

thanks a lot. I'm waiting for your help

Answer 4

you have to be patient because i get a lot of phone calls

Answer 5

sure. i have a bunch of homework for next 3 tests

Answer 6

take your time.

Answer 7

the answer is zero

Answer 8

i will explain and prove it for u but let me elaborate a bit

Answer 9

as you can see we are integrating with respect the t

Answer 10

but the limit of the the integrals x and y are integers

Answer 11

this will lead to the fact of taking the partial derivative of an integer which is zero

Answer 12

now let me prove that claim mathematically

Answer 13

not that. sorry, I don't agree with you

Answer 14

you will be patient

Answer 15

i know you what you want to say but i will show u how ur variables will be constants

Answer 16

yeah, but the answer in book is not zero, and mine is not the same with the book, that's why i need help

Answer 17

ok

Answer 18

wait are you taking the partial derivative with respect to t or x and y?

Answer 19

yes

Answer 20

u can't just say take the partial derivative like that

Answer 21

why not?

Answer 22

is it f(x,y) of f(t)

Answer 23

or f(t)

Answer 24

i need to know with respect to what?

Answer 25

no, i think you confuse. to me, take integral respect to t first, and then since it integral limit is x to y, replace those value to the result of integral, we have the function respect to x and y. at that time, we take partial derivatives for them

Answer 26

if you take the PD w r t x and y then the answer is different it won't be zero but if you take the answer with respect to t it will be zero

Answer 27

so u should kindly write the question f(x,y) otherwise the question is in complete

Answer 28

now the answer u get with respect to x is -cos(e^x)

Answer 29

Hey, friend, I think the problem is expressed on the way it is. I just copy from the book, exactly what it is

Answer 30

the answer u get with respect y is cos(e^y)

Answer 31

i will show u how

Answer 32

?

Answer 33

well if this is how the question is written in the book as u claim then throw in the bin because it is not a professional mathematical book

Answer 34

you are near right with the book, it's (respect to x) is cos (e^x) not (-) . but how? my integral is not that. please, explain

Answer 35

anyways let me show you the solution

Answer 36

I choose your recommendation to throw the book to bin is the "best response" LOL

Answer 37

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

Answer 38

where y is constant

Answer 39

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

Answer 40

this time x is constant

Answer 41

i hope this matches with ur bin book

Answer 42

hehehe

Answer 43

wait. how can you get int of cos (e^t)dt = sin (e^t)?. it must be sin(e^t) *e^t , right? because e^t is not a number, it is a function respect to t

Answer 44

I just stuck that at that part, the leftover is ok to me.

Answer 45

is the answer correct

Answer 46

so i will explain to u what is going on

Answer 47

i have a phone brb

Answer 48

\[\int\limits_{x}^{y}\cos(e^t)dt= [\sin(e^t)*e^t] from x \to y\]
replace x and y we have e^x sin(e^x) -e^y sin(e^y). right? then take derivative that part.

Answer 49

oh, it is sin (e^x)/e^x -sin(e^y)/e^y

Answer 50

i don't get. confuse everything. i'm waiting for your help.

Answer 51

i am sleepy. if you still want to explain, please let it here, i will be back tomorrow, if not, i will send you message to get the answer. thank you verymuch

Answer 52

sorry back don't be confused the function multiplied by the derivative in its chain when you integrate and differentiate u get back to where u were taking under consideration the signs, i will show u the steps tomorrow i need to go sorry about that