## University Calculus: Early Transcendentals (3rd Edition)

The tangent to $f(x)$ at $(2,1)$ is $$y=-\frac{1}{2}x+2$$
$$y=f(x)=\frac{8}{x^2+4}$$ First, we look for $f'(x)$, applying the Derivative Quotient Rule: $$f'(x)=\frac{(8)'(x^2+4)-8(x^2+4)'}{(x^2+4)^2}=\frac{0(x^2+4)-8\times2x}{(x^2+4)^2}=-\frac{16x}{(x^2+4)^2}$$ The slope of the tangent to $f(x)$ at $(2,1)$ is $$f'(2)=-\frac{16\times2}{(2^2+4)^2}=-\frac{32}{8^2}=-\frac{32}{64}=-\frac{1}{2}$$ The tangent to $f(x)$ at $(2,1)$ is $$y-1=f'(2)(x-2)$$ $$y-1=-\frac{1}{2}(x-2)$$ $$y-1=-\frac{1}{2}x+1$$ $$y=-\frac{1}{2}x+2$$