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