Question

In Exercises, use implicit differentiation to find an equation of the tangent line to the graph at the given point. $$y^2+\ln (xy)=2,(e,1)$$

Short answer

The equation of the tangent line to the curve $y^2+\ln (xy)=2$ at the point $(e,1)$ is $y=-\frac{1}{2+\frac{1}{e}}(x-e)+1$.

Step-by-step solution

Differentiate the given equation implicitly

Differentiate both sides of the equation with respect to $x$ to obtain the first derivative. Remember, when differentiating the natural logarithm of a function, the derivative is the derivative of the function divided by the function. Here both $y$ and $x$ are functions of some variable. Thus, by the Chain rule, differentiate $y^2$ to obtain $2y\frac{dy}{dx}$, and differentiate $\ln (xy)$ to obtain $\frac{1}{xy}$ times the derivative of $xy$ with respect to $x$ by the Chain Rule, producing $y+x\frac{dy}{dx}$. The derivative of $2$ is $0$. Hence, we obtain the following equation: $$2y\frac{dy}{dx}+\frac{1}{x}[y+x\frac{dy}{dx}]=0$$

Rearrange to solve for dy/dx

Rearrange our obtained equation to solve for $\frac{dy}{dx}$. This gives $\frac{dy}{dx}$(2y + $\frac{1}{x}$) = -$\frac{y}{x}$. Solving for $\frac{dy}{dx}$ yields $$\frac{dy}{dx}=-\frac{y}{2y+\frac{1}{x}}$$

Substitute the given point into the dy/dx expression

Now, to get the slope of the tangent line using the derivative above, we substitute the point $(e,1)$ into our expression for $\frac{dy}{dx}$. This yields $\frac{dy}{dx}=-\frac{1}{2(1)+\frac{1}{e}}=-\frac{1}{2+\frac{1}{e}}$

Write an equation of the tangent line at the given point

The equation of the tangent line can be represented by $$y=m(x-x1)+y1$$ where $m$ is the slope of the tangent line (which we calculated in the previous step as $\frac{dy}{dx}$), and $x1$ and $y1$ are the coordinates of the given point. Substituting the slope and the point (e, 1), we have the equation of the tangent line as $y=-\frac{1}{2+\frac{1}{e}}(x-e)+1$