Partial Derivative Calculator

Your input: find $$$\frac{\partial}{\partial y}\left(- 2 x^{2} + 2 x^{2 y} + y^{3} - 2 y^{2} + 2\right)$$$

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

$${\color{red}{\frac{\partial}{\partial y}\left(- 2 x^{2} + 2 x^{2 y} + y^{3} - 2 y^{2} + 2\right)}}={\color{red}{\left(\frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 x^{2}\right) + \frac{\partial}{\partial y}\left(2 x^{2 y}\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)\right)}}$$

Apply the constant multiple rule $$$\frac{\partial}{\partial y} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial y} \left(f \right)$$$ with $$$c=2$$$ and $$$f=x^{2 y}$$$:

$${\color{red}{\frac{\partial}{\partial y}\left(2 x^{2 y}\right)}} + \frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)={\color{red}{\left(2 \frac{\partial}{\partial y}\left(x^{2 y}\right)\right)}} + \frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)$$

Write the function $$$x^{2 y}$$$ as a composition of the two functions $$$u=g=2 y$$$ and $$$f\left(u\right)=x^{u}$$$.

Apply the chain rule $$$\frac{\partial}{\partial y} \left(f\left(g\right) \right)=\frac{\partial}{\partial u} \left(f\left(u\right) \right) \cdot \frac{\partial}{\partial y} \left(g \right)$$$:

$$2 {\color{red}{\frac{\partial}{\partial y}\left(x^{2 y}\right)}} + \frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)=2 {\color{red}{\frac{\partial}{\partial u}\left(x^{u}\right) \frac{\partial}{\partial y}\left(2 y\right)}} + \frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)$$

Apply the exponential rule $$$\frac{\partial}{\partial u} \left(a^{u} \right)=a^{u} \ln(a)$$$ with $$$a=x$$$:

$$2 {\color{red}{\frac{\partial}{\partial u}\left(x^{u}\right)}} \frac{\partial}{\partial y}\left(2 y\right) + \frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)=2 {\color{red}{x^{u} \ln{\left(x \right)}}} \frac{\partial}{\partial y}\left(2 y\right) + \frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)$$

Return to the old variable:

$$2 \ln{\left(x \right)} x^{{\color{red}{u}}} \frac{\partial}{\partial y}\left(2 y\right) + \frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)=2 \ln{\left(x \right)} x^{{\color{red}{\left(2 y\right)}}} \frac{\partial}{\partial y}\left(2 y\right) + \frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)$$

Apply the constant multiple rule $$$\frac{\partial}{\partial y} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial y} \left(f \right)$$$ with $$$c=2$$$ and $$$f=y$$$:

$$2 x^{2 y} \ln{\left(x \right)} {\color{red}{\frac{\partial}{\partial y}\left(2 y\right)}} + \frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)=2 x^{2 y} \ln{\left(x \right)} {\color{red}{\left(2 \frac{\partial}{\partial y}\left(y\right)\right)}} + \frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)$$

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

$$4 x^{2 y} \ln{\left(x \right)} {\color{red}{\frac{\partial}{\partial y}\left(y\right)}} + \frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)=4 x^{2 y} \ln{\left(x \right)} {\color{red}{1}} + \frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)$$

The derivative of a constant is 0:

$$4 x^{2 y} \ln{\left(x \right)} - {\color{red}{\frac{\partial}{\partial y}\left(2 x^{2}\right)}} + \frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)=4 x^{2 y} \ln{\left(x \right)} - {\color{red}{\left(0\right)}} + \frac{\partial}{\partial y}\left(2\right) - \frac{\partial}{\partial y}\left(2 y^{2}\right) + \frac{\partial}{\partial y}\left(y^{3}\right)$$

Apply the constant multiple rule $$$\frac{\partial}{\partial y} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial y} \left(f \right)$$$ with $$$c=2$$$ and $$$f=y^{2}$$$:

$$4 x^{2 y} \ln{\left(x \right)} - {\color{red}{\frac{\partial}{\partial y}\left(2 y^{2}\right)}} + \frac{\partial}{\partial y}\left(2\right) + \frac{\partial}{\partial y}\left(y^{3}\right)=4 x^{2 y} \ln{\left(x \right)} - {\color{red}{\left(2 \frac{\partial}{\partial y}\left(y^{2}\right)\right)}} + \frac{\partial}{\partial y}\left(2\right) + \frac{\partial}{\partial y}\left(y^{3}\right)$$

Apply the power rule $$$\frac{\partial}{\partial y} \left(y^{n} \right)=n\cdot y^{-1+n}$$$ with $$$n=2$$$:

$$4 x^{2 y} \ln{\left(x \right)} - 2 {\color{red}{\frac{\partial}{\partial y}\left(y^{2}\right)}} + \frac{\partial}{\partial y}\left(2\right) + \frac{\partial}{\partial y}\left(y^{3}\right)=4 x^{2 y} \ln{\left(x \right)} - 2 {\color{red}{\left(2 y^{-1 + 2}\right)}} + \frac{\partial}{\partial y}\left(2\right) + \frac{\partial}{\partial y}\left(y^{3}\right)=4 x^{2 y} \ln{\left(x \right)} - 4 y + \frac{\partial}{\partial y}\left(2\right) + \frac{\partial}{\partial y}\left(y^{3}\right)$$

The derivative of a constant is 0:

$$4 x^{2 y} \ln{\left(x \right)} - 4 y + {\color{red}{\frac{\partial}{\partial y}\left(2\right)}} + \frac{\partial}{\partial y}\left(y^{3}\right)=4 x^{2 y} \ln{\left(x \right)} - 4 y + {\color{red}{\left(0\right)}} + \frac{\partial}{\partial y}\left(y^{3}\right)$$

Apply the power rule $$$\frac{\partial}{\partial y} \left(y^{n} \right)=n\cdot y^{-1+n}$$$ with $$$n=3$$$:

$$4 x^{2 y} \ln{\left(x \right)} - 4 y + {\color{red}{\frac{\partial}{\partial y}\left(y^{3}\right)}}=4 x^{2 y} \ln{\left(x \right)} - 4 y + {\color{red}{\left(3 y^{-1 + 3}\right)}}=4 x^{2 y} \ln{\left(x \right)} + 3 y^{2} - 4 y$$

Thus, $$$\frac{\partial}{\partial y}\left(- 2 x^{2} + 2 x^{2 y} + y^{3} - 2 y^{2} + 2\right)=4 x^{2 y} \ln{\left(x \right)} + 3 y^{2} - 4 y$$$

Answer: $$$\frac{\partial}{\partial y}\left(- 2 x^{2} + 2 x^{2 y} + y^{3} - 2 y^{2} + 2\right)=4 x^{2 y} \ln{\left(x \right)} + 3 y^{2} - 4 y$$$