Question

If e^(xy) = 2, then what is dy/dx at the point (1, ln2)

Answer 1

we can consider that xy = ln(2)

taking the derivative we have:

x'y + xy' = 0

or: y + x(dy/dx) = 0

dy/dx = -y/x seems fair to me

Answer 2

you are trying to find the slope/equation of the tangent line, of the implicitly defined function e^(xy)=2.

~ differentiate
~ solve for dy/dx (or for y' , whatever notation you use)
~ plug in 1 for x, and ln(2) for y, into the derivative.

this will be the slope.

then use point slope form to find the euqtion of the tangent line.

Answer 3

dear @amistre64 few days ago you sent me a warning because I wrote the solution entirely, but now I see that you also have wrote the solution entirely!

Answer 4

~ have written.
~ not entirely, although to at least solve for the derivative (to isolate it), if not to take the derivative, is what the poster should have done herself.

you have a point though....

Answer 5

excuse me, * himself.

Answer 6

Thank you! @SolomonZelman

Answer 7

just expressing what I think, once we have already hit this topic. yw....

Answer 8

there is a fine line between a demonstrating how to work a problem, and doing the all the work for someone.

i do admit that there are times when i may assume too much in the way of the askers knowledge. in those cases i take it for granted that the process may not be so evident to them.

in this question, there are multiple ways to approach the solution, i simply demonstrated one of the approaches. assuming of course that they are more familiar with the implicit derivative and working the xy part of the function.

Answer 9

when a person is asking how to solve 3x + 4 = 5, i take it that they really need the hands on experience moreso than a person asking for a derivative ... mea culpa i spose.

Answer 10

Sorry I do not agree, because I think that a demonstration is already a solution!

Answer 11

then you do not agree ...

Answer 12

amistre, I think you did too much of the work for the poster in this question.

Answer 13

Please, note tah, you can consider an exercise as a simple statement in which initial data is thr hypothesis, and the final answer is the thesis

Answer 14

oops...note that...

Answer 15

also, you are forgetting about the e^(xy) itself that is there... so you have not taken the derivative correctly. e^(xy) doesn't just disappear.

Answer 16

e^(xy) = 2 is a constraint, as such, we can log both sides to form the equivalent

xy = ln(2) nothing has disappeared.

Answer 17

i was trying to think if we could approach it with a gradient, but nothing worthwhile was coming to mind :)

Answer 18

oh, I see, sorry. I was stupid not to notice that you ln-ed both sides.

Answer 19

it was most likely my handwriting, its usually not legible ;)

Answer 20

oh, it's probably because for not smart people like me, you didn't write that you took the natural log of both sides.