Question

Find an equation of the tangent line to the graph of the function at the point (1, 1).
1 + ln(xy) = e^(x-y)

Answer 1

can you help me? :)

Answer 2

a tangent line to a function is just the derivative of the line, so you need to take the derivative of your function.

d/dx(1+ln(xy)-e^(x-y)) (this is slightly rearranged from your initial function)
d/dx(1)=0 d/dx(ln(xy))=1/x and d/dx of -e^(x-y)=-e^(x-y)
so you have 1/x-e^(x-y)
then you need to evaluate it at the point (1,1) so just substitute in your point values into the appropriate variables
1/1-e^(1-1) = 1-e^0 = 1-1 = 0

Answer 3

hmmm im not sure I am following your work. Also is the equation 0?

Answer 4

yes, i just switched it over to one side to track things easier myself, sorry for not clarifying. what aren't you following?

Answer 5

do you know have to take the derivative of something?

Answer 6

but i do not believe the answer is y=0

Answer 7

yes i do, but i was confused how to take the derivative of this function

Answer 8

i could have messed something up, i'm not a math professor let me double check

Answer 9

haha thank you

Answer 10

is what you have typed how it's presented to you in the book? is there any other information?

Answer 11

it is an online homework thing called webassign and that is how the problem was presented

Answer 12

hmm we use webassign as well unfortunately. i can't really find much in my old notes regarding setups like this unfortunately. hopefully someone else can help you out.

Answer 13

alright then, thank you for trying

Answer 14

yeah, sorry i couldn't help.