Калькулятор частинних похідних

Your input: find $\frac{\partial}{\partial y}\left(3 x + 4 y\right)$

The derivative of a sum/difference is the sum/difference of derivatives:

$${\color{red}{\frac{\partial}{\partial y}\left(3 x + 4 y\right)}}={\color{red}{\left(\frac{\partial}{\partial y}\left(3 x\right) + \frac{\partial}{\partial y}\left(4 y\right)\right)}}$$

Apply the constant multiple rule $\frac{\partial}{\partial y} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial y} \left(f \right)$ with $c=4$ and $f=y$:

$${\color{red}{\frac{\partial}{\partial y}\left(4 y\right)}} + \frac{\partial}{\partial y}\left(3 x\right)={\color{red}{\left(4 \frac{\partial}{\partial y}\left(y\right)\right)}} + \frac{\partial}{\partial y}\left(3 x\right)$$

Apply the power rule $\frac{\partial}{\partial y} \left(y^{n} \right)=n\cdot y^{-1+n}$ with $n=1$, in other words $\frac{\partial}{\partial y} \left(y \right)=1$:

$$4 {\color{red}{\frac{\partial}{\partial y}\left(y\right)}} + \frac{\partial}{\partial y}\left(3 x\right)=4 {\color{red}{1}} + \frac{\partial}{\partial y}\left(3 x\right)$$

The derivative of a constant is 0:

$$4 + {\color{red}{\frac{\partial}{\partial y}\left(3 x\right)}}=4 + {\color{red}{\left(0\right)}}$$

Thus, $\frac{\partial}{\partial y}\left(3 x + 4 y\right)=4$

Answer: $\frac{\partial}{\partial y}\left(3 x + 4 y\right)=4$