Question

differentiate the following: ((cos(x)(e^x-e^a))/(x-a)(e^x))-((sin(x)*ln(x-a)(e^x-e^a))/(e^x(x-a))

Answer 1

haha ... use wolfram :)

Answer 2

also you need to differentiate numerator and denominator separately

Answer 3

I did use Wolfram and boy, it disagreed with both you and I. I'm very, very lost now.

Answer 4

here you go
http://www.wolframalpha.com/input/?i=differentiate+cos%28x%29*%28e^x-e^a%29-sin%28x%29*%28x-a%29*ln%28x-a%29%28e^x-e^a%29

now evaluate x=a

Answer 5

No, I hadn't gotten that far yet.

Answer 6

I'd moved on to another impossible problem. Anyway, backtracking now.

Answer 7

so what I saw on Wolfram had no denominator, that because you're differentiating the denominator from the numerator seperately and you just sent me the numerator?

Answer 8

yes correct...the denominator should be easier

Answer 9

wow, this just reeks of IMPOSSIBLE

Answer 10

how do I take that monstrosity and get an appreciable answer?

Answer 11

the result in Wolfram? plug in x=a and see what it simplifies to
..should get a bunch of zeros

Answer 12

yeah, well okay, I can see there's a lot of x-a's. So my numerator is essentially e^xcos(x) would you say?

Answer 13

correct ... and the denominator is ?

Answer 14

hmmm, it looks like e^x(x-a)

Answer 15

right? (u)e^x *(1), u = (x-a)?

Answer 16

f'(x) of e^x is e^x, right?

Answer 17

so much to remember algebraically I forget all the smooth and easy stuff

Answer 18

yep ...du = 1

Answer 19

great, so I'm left with (e^x (cos x))/e^x(x-a)

Answer 20

the e^x's = 1, so I'm finished with cos x/(x-a)?

Answer 21

x-a = 0, so I have to differentiate a 3rd time?

Answer 22

cos X/0

Answer 23

not quite ... product rule says you add

--> e^x + e^x(x-a)

Answer 24

oh, it's so easy to forget the simple stuff after fighting with all these SYMBOLS for so long. What have us humans come to?

Answer 25

haha :)

Answer 26

so after applying limit ...x=a

--> e^a cos(a) / e^a = cos(a)

Answer 27

whoa! Not what I am seeing. I have (before replacing x) (e^x(Cosx))/(2e^x(x-a))

Answer 28

maybe I did that wrong, maybe the denominator is (e^x(x-a))+e^x?

Answer 29

yes thats it....you can't combine them to make 2e^x

Answer 30

Perrrrrrfect! You are uber patient.

Answer 31

Cos a is my answer then!

Answer 32

alright good you see it now ...yep