Question

1. (6 points) consider the curve given by the equation: $ln(xy)=2x$.

a. (4 points) find $\frac{dy}{dx}$
b. (2 points) find the tangent line to the curve at $(1,e^{2})$.

Explanation:

Step1: Differentiate both sides of the equation

Differentiate $\ln(xy)$ and $2x$ with respect to $x$.
Using the chain rule and product rule, $\frac{d}{dx}(\ln(xy))=\frac{1}{xy}(y + x\frac{dy}{dx})$ and $\frac{d}{dx}(2x)=2$.
So, $\frac{1}{xy}(y + x\frac{dy}{dx})=2$.

Step2: Solve for $\frac{dy}{dx}$

Multiply both sides by $xy$: $y + x\frac{dy}{dx}=2xy$.
Then, $x\frac{dy}{dx}=2xy - y$.
Factor out $y$: $x\frac{dy}{dx}=y(2x - 1)$.
Finally, $\frac{dy}{dx}=\frac{y(2x - 1)}{x}$.

Step3: Find the slope of the tangent line at $(1,e^{2})$

Substitute $x = 1$ and $y = e^{2}$ into $\frac{dy}{dx}$: $\frac{dy}{dx}\vert_{x = 1,y = e^{2}}=\frac{e^{2}(2\times1 - 1)}{1}=e^{2}$.

Step4: Use the point - slope form to find the tangent line

The point - slope form is $y - y_{0}=m(x - x_{0})$, where $(x_{0},y_{0})=(1,e^{2})$ and $m = e^{2}$.
So, $y - e^{2}=e^{2}(x - 1)$.
Simplify: $y - e^{2}=e^{2}x - e^{2}$, which gives $y = e^{2}x$.

Answer:

a. $\frac{dy}{dx}=\frac{y(2x - 1)}{x}$
b. $y = e^{2}x$