Question

Finding an Equation of a Tangent Line In Exercise, use implicit differentiation to find an equation of the tangent line to the graph of the function at the given point.
y2 + ln(xy) = 2; (e, 1)

Answer 1

Answer:

$y=\frac{4}{3}-\frac{x}{3e}$

Step-by-step explanation:

Equation of tangent line of given function at (e,1) is found by implicitly differentiating the given function w.r.to x

$y^{2}+ln(xy)=2\\\\y^{2}+ln(x)+ln(y)=2\\\\\frac{d}{dx}(y^{2}+ln(x)+ln(y))=\frac{d}{dx}(2)\\\\2y\frac{dy}{dx}+\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=0\\\\\frac{dy}{dx}(2y+\frac{1}{y})=-\frac{1}{x}\\\\\frac{dy}{dx}=-\frac{y}{x(2y^{2}+1)}\\\\at (e,1)\\\\\frac{dy}{dx}=-\frac{1}{e(2+1)}\\\\\frac{dy}{dx}=\frac{1}{3e}$

Equation of tangent line is

y-y₁=m(x-x₁)

at (e,1)

$y-1=-\frac{1}{3e}(x-e)\\\\y=1-\frac{1}{3e}(x-e)\\\\y=1+\frac{1}{3}-\frac{x}{3e}\\\\y=\frac{4}{3}-\frac{x}{3e}$