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

The tangent line to the curve at $x=0$ is $y=-4x+1$.
$$y=f(x)=\Big(\frac{x-1}{x+1}\Big)^2$$ a) Find $f'(x)$: Using the Chain Rule: $$f'(x)=2\Big(\frac{x-1}{x+1}\Big)\Big(\frac{x-1}{x+1}\Big)'=2\Big(\frac{x-1}{x+1}\Big)\frac{(x-1)'(x+1)-(x-1)(x+1)'}{(x+1)^2}$$ $$f'(x)=2\Big(\frac{x-1}{x+1}\Big)\frac{(x+1)-(x-1)}{(x+1)^2}=2\Big(\frac{x-1}{x+1}\Big)\frac{2}{(x+1)^2}$$ $$f'(x)=\frac{4(x-1)}{(x+1)^3}$$ b) For $x=0$, we have $$f(0)=\Big(\frac{0-1}{0+1}\Big)^2=\Big(\frac{-1}{1}\Big)^2=1$$ $$f'(0)=\frac{4(0-1)}{(0+1)^3}=\frac{-4}{1^3}=-4$$ As $f'(0)$ is the slope of the tangent line to $f(x)$ at $x=0$, we can find the tangent to $f(x)$ at $x=0$ following the formula: $$y-f(0)=f'(0)(x-0)$$ $$y-1=-4x$$ $$y=-4x+1$$