Respuesta :
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
[tex]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\\\\[/tex]
which is slope of tangent line. At (1,0) equation of tangent line is
[tex]y-0=1(x-1)\\\\y=x-1[/tex]
