Exponential Function: for xe^y + ye^x = 2e, P(1,1), find...

[rich gill]

Hi All,
I am new to the site, a friend recommended it to me.
I have been struggling through Calc 1 and reading this site has helped alot, anyhow I have 2 problems that I dont even know how to start.

1)
Given: xe^y + ye^x = 2e
P(1,1)
Find dy @ P

2)
Given: e^xy = e^2
P(2,1)
Find dy @ P, if dx = 6


Thanks for any help ahead of time

Rich

 

[galactus]

Re: The Exponential Function

If I understand correctly, you can use implicit differentiation.

\(\displaystyle xe^{y}+ye^{x}=2e\)

Product rule on each term:

\(\displaystyle xe^{y}\frac{dy}{dx}+e^{y}+ye^{x}+e^{x}\frac{dy}{dx}=0\)

\(\displaystyle \frac{dy}{dx}\left(xe^{y}+e^{x}\right)=-e^{y}-ye^{x}\)

\(\displaystyle \boxed{\frac{dy}{dx}=\frac{-(ye^{x}+e^{y})}{xe^{y}+e^{x}}}\)

Now, sub in your values for x and y to find the value at P.

Now, you try ID on the other one.

 

[rich gill]

Thanks for the help, I did the second problem however it comes out just like the first, and I am not sure where to sub in my DX value?

Thanks again


Rich