Answer
$$R'(0)=1$$
Work Step by Step
$$R(x)=\frac{x-3x^3+5x^5}{1+3x^3+6x^6+9x^9}$$ $$R(x)=\frac{f(x)}{g(x)}$$
According to Quotient Rule, $$R'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$$ Therefore, $$R'(0)=\frac{f'(0)g(0)-f(0)g'(0)}{[g(0)]^2}$$
*Find $f(0)$, $g(0)$, $f'(0)$ and $g'(0)$
$$f(x)=x-3x^3+5x^5$$
So, $$f'(x)=1-9x^2+25x^4$$
Therefore, $f(0)=0-3\times0^3+5\times0^5=0$
and $f'(0)=1-9\times0^2+25\times0^4=1$
$$g(x)=1+3x^3+6x^6+9x^9$$
So, $$g'(x)=9x^2+36x^5+81x^8$$
Therefore, $g(0)=1+3\times0^3+6\times0^6+9\times0^9=1$
and $g'(0)=9\times0^2+36\times0^5+81\times0^8=0$
That means $$R'(0)=\frac{1\times1-0\times0}{1^2}=1$$
