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