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

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

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

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

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

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

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

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

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

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

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

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

The derivative of a constant is 0:

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

The derivative of a constant is 0:

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

The derivative of a constant is 0:

$$4 x y - 4 x + {\color{red}{\frac{\partial}{\partial x}\left(y^{3}\right)}}=4 x y - 4 x + {\color{red}{\left(0\right)}}=4 x \left(y - 1\right)$$

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

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

Apply the constant multiple rule $$$\frac{\partial}{\partial x} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial x} \left(f \right)$$$ with $$$c=4 \left(y - 1\right)$$$ and $$$f=x$$$:

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

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

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

Thus, $$$\frac{\partial}{\partial x}\left(4 x \left(y - 1\right)\right)=4 y - 4$$$

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

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