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

`y=(4)/(3)-(x)/(3e)`

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\\\\(d)/(dx)(y^(2)+ln(x)+ln(y))=(d)/(dx)(2)\\\\2y(dy)/(dx)+(1)/(x)+(1)/(y)(dy)/(dx)=0\\\\(dy)/(dx)(2y+(1)/(y))=-(1)/(x)\\\\(dy)/(dx)=-(y)/(x(2y^(2)+1))\\\\at (e,1)\\\\(dy)/(dx)=-(1)/(e(2+1))\\\\(dy)/(dx)=(1)/(3e)`

Equation of tangent line is

y-y₁=m(x-x₁)

at (e,1)

`y-1=-(1)/(3e)(x-e)\\\\y=1-(1)/(3e)(x-e)\\\\y=1+(1)/(3)-(x)/(3e)\\\\y=(4)/(3)-(x)/(3e)\\`