Answer
The tangent to $f(x)$ at the origin is $y=4x$. The tangent to $f(x)$ at $(1,2)$ is $y=2$
Work Step by Step
$$y=f(x)=\frac{4x}{x^2+1}$$
1) First, we look for $f'(x)$, applying the Derivative Quotient Rule: $$f'(x)=\frac{(4x)'(x^2+1)-4x(x^2+1)'}{(x^2+1)^2}=\frac{4(x^2+1)-4x\times2x}{(x^2+1)^2}$$
$$f'(x)=\frac{4x^2+4-8x^2}{(x^2+1)^2}=\frac{4-4x^2}{(x^2+1)^2}$$
2) At the origin: $$f'(0)=\frac{4-4\times0^2}{(0^2+1)^2}=\frac{4-0}{1}=4$$
The tangent to $f(x)$ at the origin is $$y-0=f'(0)(x-0)$$
$$y=f'(0)x=4x$$
3) At $(1,2)$: $$f'(1)=\frac{4-4\times1^2}{(1^2+1)^2}=\frac{4-4}{2^2}=\frac{0}{4}=0$$
The tangent to $f(x)$ at $(1,2)$ is $$y-2=f'(1)(x-1)=0(x-1)$$
$$y-2=0$$
$$y=2$$
