Calculus: Early Transcendentals 8th Edition

The equation of the tangent line $$(l): y=1$$ The equation of the normal line $$(m): x=1$$
$$y=f(x)=\frac{2x}{x^2+1}$$ 1) Find $f'(x)$ $$f'(x)=\frac{(2x)'(x^2+1)-2x(x^2+1)'}{(x^2+1)^2}$$ $$f'(x)=\frac{2(x^2+1)-2x\times2x}{(x^2+1)^2}$$ $$f'(x)=\frac{2x^2+2-4x^2}{(x^2+1)^2}$$ $$f'(x)=\frac{2-2x^2}{(x^2+1)^2}$$ 2) The slope of the tangent line $l$ to the given curve at point $A(a,b)$ is $f'(a)$. Therefore, at $(1,1)$, the slope of tangent line $l$ is $$f'(1)=\frac{2-2\times1^2}{(1^2+1)^2}$$ $$f'(1)=\frac{2-2}{2^2}=0$$ The equation of the tangent line $l$ to the given curve at $(1,1)$ has the following form: $$(l): (y-1)=f'(1)(x-1)$$ $$(l): y-1=0(x-1)$$ $$(l): y=1$$ 3) The normal line $m$ is perpendicular with the tangent line $l$ and also passes through $(1,1)$. Therefore, since the tangent line $l$ has equation $y=1$, the normal line $m$ would have equation $$(m): x=1$$