## Calculus: Early Transcendentals 8th Edition

$$Q'(0)=4$$
$$Q(x)=\frac{1+x+x^2+xe^x}{1-x+x^2-xe^x}$$ $$R(x)=\frac{f(x)}{g(x)}$$ According to Quotient Rule, $$Q'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$$ Therefore, $$Q'(0)=\frac{f'(0)g(0)-f(0)g'(0)}{[g(0)]^2}$$ *Find $f(0)$, $g(0)$, $f'(0)$ and $g'(0)$ $$f(x)=1+x+x^2+xe^x$$ So, $$f'(x)=1+2x+e^x+xe^x$$ Therefore, $f(0)=1+0+0^2+0\times e^0=1$ and $f'(0)=1+2\times0+e^0+0\times e^0=2$ $$g(x)=1-x+x^2-xe^x$$ So, $$g'(x)=-1+2x-e^x-xe^x$$ Therefore, $g(0)=1-0+0^2-0\times e^0=1$ and $g'(0)=-1+2\times0-e^0-0\times e^0=-2$ That means $$Q'(0)=\frac{2\times1-1\times(-2)}{1^2}=4$$