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Solution
Your input: find $$$\frac{\partial}{\partial x}\left(4 x + y\right)$$$
The derivative of a sum/difference is the sum/difference of derivatives:
$${\color{red}{\frac{\partial}{\partial x}\left(4 x + y\right)}}={\color{red}{\left(\frac{\partial}{\partial x}\left(4 x\right) + \frac{\partial}{\partial x}\left(y\right)\right)}}$$Apply the constant multiple rule $$$\frac{\partial}{\partial x} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial x} \left(f \right)$$$ with $$$c=4$$$ and $$$f=x$$$:
$${\color{red}{\frac{\partial}{\partial x}\left(4 x\right)}} + \frac{\partial}{\partial x}\left(y\right)={\color{red}{\left(4 \frac{\partial}{\partial x}\left(x\right)\right)}} + \frac{\partial}{\partial x}\left(y\right)$$Apply the power rule $$$\frac{\partial}{\partial x} \left(x^{n} \right)=n\cdot x^{-1+n}$$$ with $$$n=1$$$, in other words $$$\frac{\partial}{\partial x} \left(x \right)=1$$$:
$$4 {\color{red}{\frac{\partial}{\partial x}\left(x\right)}} + \frac{\partial}{\partial x}\left(y\right)=4 {\color{red}{1}} + \frac{\partial}{\partial x}\left(y\right)$$The derivative of a constant is 0:
$$4 + {\color{red}{\frac{\partial}{\partial x}\left(y\right)}}=4 + {\color{red}{\left(0\right)}}$$Thus, $$$\frac{\partial}{\partial x}\left(4 x + y\right)=4$$$
Answer: $$$\frac{\partial}{\partial x}\left(4 x + y\right)=4$$$
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