Derivative of $$$- \sqrt{3} x + \sqrt{2} x$$$
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Your Input
Find $$$\frac{d}{dx} \left(- \sqrt{3} x + \sqrt{2} x\right)$$$.
Solution
The derivative of a sum/difference is the sum/difference of derivatives:
$${\color{red}\left(\frac{d}{dx} \left(- \sqrt{3} x + \sqrt{2} x\right)\right)} = {\color{red}\left(- \frac{d}{dx} \left(\sqrt{3} x\right) + \frac{d}{dx} \left(\sqrt{2} x\right)\right)}$$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 $$$c = \sqrt{3}$$$ and $$$f{\left(x \right)} = x$$$:
$$- {\color{red}\left(\frac{d}{dx} \left(\sqrt{3} x\right)\right)} + \frac{d}{dx} \left(\sqrt{2} x\right) = - {\color{red}\left(\sqrt{3} \frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(\sqrt{2} x\right)$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$- \sqrt{3} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(\sqrt{2} x\right) = - \sqrt{3} {\color{red}\left(1\right)} + \frac{d}{dx} \left(\sqrt{2} x\right)$$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 $$$c = \sqrt{2}$$$ and $$$f{\left(x \right)} = x$$$:
$${\color{red}\left(\frac{d}{dx} \left(\sqrt{2} x\right)\right)} - \sqrt{3} = {\color{red}\left(\sqrt{2} \frac{d}{dx} \left(x\right)\right)} - \sqrt{3}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$\sqrt{2} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} - \sqrt{3} = \sqrt{2} {\color{red}\left(1\right)} - \sqrt{3}$$Thus, $$$\frac{d}{dx} \left(- \sqrt{3} x + \sqrt{2} x\right) = - \sqrt{3} + \sqrt{2}$$$.
Answer
$$$\frac{d}{dx} \left(- \sqrt{3} x + \sqrt{2} x\right) = - \sqrt{3} + \sqrt{2}$$$A