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Partial Derivative Calculator

Calculate partial derivatives step by step

This online calculator will calculate the partial derivative of the function, with steps shown. You can specify any order of integration.

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Hint: type x^2,y to calculate `(partial^3 f)/(partial x^2 partial y)`, or enter x,y^2,x to find `(partial^4 f)/(partial x partial y^2 partial x)`.

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Solution

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

First, find $$$\frac{\partial}{\partial y}\left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right)$$$

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

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

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

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

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

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

The derivative of a constant is 0:

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

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 $$$f=y^{2}$$$:

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

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

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

The derivative of a constant is 0:

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

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

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

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

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

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

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

Apply the constant multiple rule $$$\frac{\partial}{\partial y} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial y} \left(f \right)$$$ with $$$c=3$$$ and $$$f=y^{2}$$$:

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

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

$$3 {\color{red}{\frac{\partial}{\partial y}\left(y^{2}\right)}} + \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(4 y\right)=3 {\color{red}{\left(2 y^{-1 + 2}\right)}} + \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(4 y\right)=6 y + \frac{\partial}{\partial y}\left(2 x^{2}\right) - \frac{\partial}{\partial y}\left(4 y\right)$$

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 $$$f=y$$$:

$$6 y - {\color{red}{\frac{\partial}{\partial y}\left(4 y\right)}} + \frac{\partial}{\partial y}\left(2 x^{2}\right)=6 y - {\color{red}{\left(4 \frac{\partial}{\partial y}\left(y\right)\right)}} + \frac{\partial}{\partial y}\left(2 x^{2}\right)$$

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

$$6 y - 4 {\color{red}{\frac{\partial}{\partial y}\left(y\right)}} + \frac{\partial}{\partial y}\left(2 x^{2}\right)=6 y - 4 {\color{red}{1}} + \frac{\partial}{\partial y}\left(2 x^{2}\right)$$

The derivative of a constant is 0:

$$6 y - 4 + {\color{red}{\frac{\partial}{\partial y}\left(2 x^{2}\right)}}=6 y - 4 + {\color{red}{\left(0\right)}}$$

Thus, $$$\frac{\partial}{\partial y}\left(2 x^{2} + 3 y^{2} - 4 y\right)=6 y - 4$$$

Therefore, $$$\frac{\partial^{2}}{\partial y^{2}}\left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right)=6 y - 4$$$

Answer: $$$\frac{\partial^{2}}{\partial y^{2}}\left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right)=6 y - 4$$$