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Differentiation rules practice question 18

Two students find f′(1)f'(1) for f(x)=2x+3x2+1f(x) = \dfrac{2x + 3}{x^{2} + 1}, as shown. Which solution is right, and why?

Solution AA1:u=2x+3, u′=2;v=x2+1, v′=2xA2:f′(x)=uv′−u′vv2=(2x+3)(2x)−2(x2+1)(x2+1)2A3:f′(1)=(5)(2)−2(2)22=64=32Solution BB1:u=2x+3, u′=2;v=x2+1, v′=2xB2:f′(x)=u′v−uv′v2=2(x2+1)−(2x+3)(2x)(x2+1)2B3:=−2x2−6x+2(x2+1)2B4:f′(1)=−2−6+24=−32\begin{aligned} &\textbf{Solution A} \\ \text{A1:}\quad & u = 2x + 3,\ u' = 2; \\ & v = x^{2} + 1,\ v' = 2x \\[4pt] \text{A2:}\quad & f'(x) = \frac{uv' - u'v}{v^{2}} \\ & = \frac{(2x + 3)(2x) - 2(x^{2} + 1)}{(x^{2} + 1)^{2}} \\[4pt] \text{A3:}\quad & f'(1) = \frac{(5)(2) - 2(2)}{2^{2}} = \frac{6}{4} = \frac{3}{2} \\[14pt] &\textbf{Solution B} \\ \text{B1:}\quad & u = 2x + 3,\ u' = 2; \\ & v = x^{2} + 1,\ v' = 2x \\[4pt] \text{B2:}\quad & f'(x) = \frac{u'v - uv'}{v^{2}} \\ & = \frac{2(x^{2} + 1) - (2x + 3)(2x)}{(x^{2} + 1)^{2}} \\[4pt] \text{B3:}\quad & = \frac{-2x^{2} - 6x + 2}{(x^{2} + 1)^{2}} \\[4pt] \text{B4:}\quad & f'(1) = \frac{-2 - 6 + 2}{4} = -\frac{3}{2} \end{aligned}

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