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 $\displaystyle \frac{dy}{dx}$.