## Calculus, 10th Edition (Anton)

f'(x)=$\frac{-3x^{2}-8x+3}{(x^{2}+1)^{2}}$
f(x)=$\frac{3x+4}{x^{2}+1}$ According to the quotient rule, if f(x)=$\frac{h(x)}{g(x)}$, then f'(x)=$\frac{g(x)h'(x)-h(x)g'(x)}{(g(x))^{2}}$. If we set h(x)=3x+4, and g(x)=$x^{2}$+1, then f'(x)=$\frac{(x^{2}+1)\frac{d}{dx}[3x+4]-(3x+4)\frac{d}{dx}[x^{2}+1]}{(x^{2}+1)^{2}}$ $\frac{d}{dx}$[3x+4]=3, using the power rule $\frac{d}{dx}$[$x^{2}$+1]=2x, using the power rule Therefore, f'(x)=$\frac{(x^{2}+1)(3)-(3x+4)(2x)}{(x^{2}+1)^{2}}$ =$\frac{(3x^{2}+3)-(6x^{2}+8x)}{(x^{2}+1)^{2}}$ =$\frac{-3x^{2}-8x+3}{(x^{2}+1)^{2}}$