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.
x + y - 1 = ln(x2 + y2); (1, 0)

Answer 1

Answer:

Equation of tangent line at point (1,0) to given function is

y=x-1

Step-by-step explanation:

Equation of tangent line is

                                   y-y₁=m(x-x₁)-----(1)

where m is slope of line and can be found by implicitly differentiating the given function w.r.to x

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

which is slope of tangent line. At (1,0) equation of tangent line is

                            $y-0=1(x-1)\\\\y=x-1$