Implicit Function differentiation

Question

For the implicit function e^x + e^y = e^x+y show that dy/dx = -e^y-x , i have no idea how to get that. ive spent an hour on the problem derivative after another cant get it can you help?

Answer 1

Starting with e^x + e^y = e^x+y

e^x = e^x+y - e^y ,subtracting e^y from both sides

e^x = e^y (e^x - 1), factoring out the e^y   Let's label this as equation (1) since we'll use it below

 

So e^y = e^x/ (e^x - 1) , dividing both sides by e^x - 1

and y = ln [e^x/(e^x - 1)], taking natural logs of both sides

      y = ln e^x - ln(e^x - 1), using natural log quotient rule

      y = x - ln(e^x - 1), simplifying since ln(e^x) = x

 

Now take the derivative with respect to x

 

dy/dx = 1 - [(1/(e^x - 1)) e^x], since f(x) = e^x - 1 and derivative of ln(f(x)) is (1/f(x)) * f'(x)

dy/dx = ((e^x - 1) - [e^x]) / (e^x - 1), by rewriting 1 as (e^x - 1)/(e^x - 1) and putting terms over common denominator

dy/dx = -1/(e^x - 1)   Let's label this as equation (2)

 

Using equation (1) above, e^x = e^y (e^x - 1)

so e^x / e^y = e^x - 1, dividing both sides by e^y

so e^x-y = e^x - 1  Let's label this as equation (3)

 

Using equation (3) to substitute for e^x - 1 in equation (2), we get:

dy/dx = -1/(e^x-y)

          = -1(e^y-x) , since e^-z =1/e^z with z = x - y

          = -e^y-x ,which is the desired result