1. ## Logarithmic Differentiation Question

I am having trouble with this questoin...

Find the slope (exact value) of the normal to the curve xe^y + ylnx = 2 at the point (1, ln2).

Thanks for any help!

2. Originally Posted by ty2391
I am having trouble with this questoin...

Find the slope (exact value) of the normal to the curve xe^y + ylnx = 2 at the point (1, ln2).

Thanks for any help!
Differentiate both sides of $\displaystyle x e^y + y \ln x = 2$ with respect to x, using the product rule as required:

$\displaystyle e^y + x \left( e^y \frac{dy}{dx} \right) + \frac{dy}{dx} \ln x + \frac{y}{x} = 0$.

Note: From the chain rule, the derivative with respect to x of $\displaystyle e^y$ is $\displaystyle \frac{d}{dy}\left[ e^y \right] \frac{dy}{dx} = e^y \frac{dy}{dx}$.

Now substitute x = 1 and y = ln 2. Note: $\displaystyle e^{\ln 2} = 2$.

Now solve for $\displaystyle \frac{dy}{dx}$.

3. I ended up with:

2 / ln 2 + 2

Can someone confirm this?

4. Originally Posted by ty2391
I ended up with:

2 / ln 2 + 2

Can someone confirm this?
$\displaystyle e^y + x \left( e^y \frac{dy}{dx} \right) + \frac{dy}{dx} \ln x + \frac{y}{x} = 0$.

Substitute x = 1 and y = ln 2:

$\displaystyle 2 + 2 \frac{dy}{dx} + \ln 2 = 0 \Rightarrow \frac{dy}{dx} = -\frac{(2 + \ln 2)}{2}$ ....

5. It's looking for the normal though, so I guess I am right?

6. Originally Posted by ty2391
It's looking for the normal though, so I guess I am right?
(My boldface). After what I posted, you need to guess!?