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

First, find $$$\frac{\partial}{\partial x}\left(x^{4} - 4 x y + y^{4} + 1\right)$$$

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

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

The derivative of a constant is 0:

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

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

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

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

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

The derivative of a constant is 0:

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

Apply the power rule $$$\frac{\partial}{\partial x} \left(x^{n} \right)=n\cdot x^{-1+n}$$$ with $$$n=4$$$:

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

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

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

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 $$$f=x^{3} - y$$$:

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

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

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

The derivative of a constant is 0:

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

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

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

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

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

Answer: $$$\frac{\partial^{2}}{\partial x \partial y}\left(x^{4} - 4 x y + y^{4} + 1\right)=-4$$$