Derivative of $$$\frac{x^{3} - 2 x^{2}}{x^{2} + 1}$$$
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Find $$$\frac{d}{dx} \left(\frac{x^{3} - 2 x^{2}}{x^{2} + 1}\right)$$$.
Solution
Apply the quotient rule $$$\frac{d}{dx} \left(\frac{f{\left(x \right)}}{g{\left(x \right)}}\right) = \frac{\frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} - f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)}{g^{2}{\left(x \right)}}$$$ with $$$f{\left(x \right)} = x^{3} - 2 x^{2}$$$ and $$$g{\left(x \right)} = x^{2} + 1$$$:
$${\color{red}\left(\frac{d}{dx} \left(\frac{x^{3} - 2 x^{2}}{x^{2} + 1}\right)\right)} = {\color{red}\left(\frac{\frac{d}{dx} \left(x^{3} - 2 x^{2}\right) \left(x^{2} + 1\right) - \left(x^{3} - 2 x^{2}\right) \frac{d}{dx} \left(x^{2} + 1\right)}{\left(x^{2} + 1\right)^{2}}\right)}$$The derivative of a sum/difference is the sum/difference of derivatives:
$$\frac{\left(x^{2} + 1\right) \frac{d}{dx} \left(x^{3} - 2 x^{2}\right) - \left(x^{3} - 2 x^{2}\right) {\color{red}\left(\frac{d}{dx} \left(x^{2} + 1\right)\right)}}{\left(x^{2} + 1\right)^{2}} = \frac{\left(x^{2} + 1\right) \frac{d}{dx} \left(x^{3} - 2 x^{2}\right) - \left(x^{3} - 2 x^{2}\right) {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(1\right)\right)}}{\left(x^{2} + 1\right)^{2}}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 2$$$:
$$\frac{\left(x^{2} + 1\right) \frac{d}{dx} \left(x^{3} - 2 x^{2}\right) - \left(x^{3} - 2 x^{2}\right) \left({\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(1\right)\right)}{\left(x^{2} + 1\right)^{2}} = \frac{\left(x^{2} + 1\right) \frac{d}{dx} \left(x^{3} - 2 x^{2}\right) - \left(x^{3} - 2 x^{2}\right) \left({\color{red}\left(2 x\right)} + \frac{d}{dx} \left(1\right)\right)}{\left(x^{2} + 1\right)^{2}}$$The derivative of a constant is $$$0$$$:
$$\frac{- \left(2 x + {\color{red}\left(\frac{d}{dx} \left(1\right)\right)}\right) \left(x^{3} - 2 x^{2}\right) + \left(x^{2} + 1\right) \frac{d}{dx} \left(x^{3} - 2 x^{2}\right)}{\left(x^{2} + 1\right)^{2}} = \frac{- \left(2 x + {\color{red}\left(0\right)}\right) \left(x^{3} - 2 x^{2}\right) + \left(x^{2} + 1\right) \frac{d}{dx} \left(x^{3} - 2 x^{2}\right)}{\left(x^{2} + 1\right)^{2}}$$The derivative of a sum/difference is the sum/difference of derivatives:
$$\frac{- 2 x \left(x^{3} - 2 x^{2}\right) + \left(x^{2} + 1\right) {\color{red}\left(\frac{d}{dx} \left(x^{3} - 2 x^{2}\right)\right)}}{\left(x^{2} + 1\right)^{2}} = \frac{- 2 x \left(x^{3} - 2 x^{2}\right) + \left(x^{2} + 1\right) {\color{red}\left(\frac{d}{dx} \left(x^{3}\right) - \frac{d}{dx} \left(2 x^{2}\right)\right)}}{\left(x^{2} + 1\right)^{2}}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 3$$$:
$$\frac{- 2 x \left(x^{3} - 2 x^{2}\right) + \left(x^{2} + 1\right) \left({\color{red}\left(\frac{d}{dx} \left(x^{3}\right)\right)} - \frac{d}{dx} \left(2 x^{2}\right)\right)}{\left(x^{2} + 1\right)^{2}} = \frac{- 2 x \left(x^{3} - 2 x^{2}\right) + \left(x^{2} + 1\right) \left({\color{red}\left(3 x^{2}\right)} - \frac{d}{dx} \left(2 x^{2}\right)\right)}{\left(x^{2} + 1\right)^{2}}$$Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = 2$$$ and $$$f{\left(x \right)} = x^{2}$$$:
$$\frac{- 2 x \left(x^{3} - 2 x^{2}\right) + \left(x^{2} + 1\right) \left(3 x^{2} - {\color{red}\left(\frac{d}{dx} \left(2 x^{2}\right)\right)}\right)}{\left(x^{2} + 1\right)^{2}} = \frac{- 2 x \left(x^{3} - 2 x^{2}\right) + \left(x^{2} + 1\right) \left(3 x^{2} - {\color{red}\left(2 \frac{d}{dx} \left(x^{2}\right)\right)}\right)}{\left(x^{2} + 1\right)^{2}}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 2$$$:
$$\frac{- 2 x \left(x^{3} - 2 x^{2}\right) + \left(x^{2} + 1\right) \left(3 x^{2} - 2 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)}\right)}{\left(x^{2} + 1\right)^{2}} = \frac{- 2 x \left(x^{3} - 2 x^{2}\right) + \left(x^{2} + 1\right) \left(3 x^{2} - 2 {\color{red}\left(2 x\right)}\right)}{\left(x^{2} + 1\right)^{2}}$$Simplify:
$$\frac{- 2 x \left(x^{3} - 2 x^{2}\right) + \left(x^{2} + 1\right) \left(3 x^{2} - 4 x\right)}{\left(x^{2} + 1\right)^{2}} = \frac{x \left(x - 1\right) \left(x^{2} + x + 4\right)}{\left(x^{2} + 1\right)^{2}}$$Thus, $$$\frac{d}{dx} \left(\frac{x^{3} - 2 x^{2}}{x^{2} + 1}\right) = \frac{x \left(x - 1\right) \left(x^{2} + x + 4\right)}{\left(x^{2} + 1\right)^{2}}$$$.
Answer
$$$\frac{d}{dx} \left(\frac{x^{3} - 2 x^{2}}{x^{2} + 1}\right) = \frac{x \left(x - 1\right) \left(x^{2} + x + 4\right)}{\left(x^{2} + 1\right)^{2}}$$$A