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.

Enter a function:

Enter the order of integration:

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)`.

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.

Solution

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

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

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

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

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

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

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

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

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

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

The derivative of a constant is 0:

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

The derivative of a constant is 0:

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

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

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

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

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

Apply the constant multiple rule $$$\frac{\partial}{\partial y} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial y} \left(f \right)$$$ with $$$c=4$$$ and $$$f=y^{3}$$$:

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

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

$$4 {\color{red}{\frac{\partial}{\partial y}\left(y^{3}\right)}} - \frac{\partial}{\partial y}\left(4 x\right)=4 {\color{red}{\left(3 y^{-1 + 3}\right)}} - \frac{\partial}{\partial y}\left(4 x\right)=12 y^{2} - \frac{\partial}{\partial y}\left(4 x\right)$$

The derivative of a constant is 0:

$$12 y^{2} - {\color{red}{\frac{\partial}{\partial y}\left(4 x\right)}}=12 y^{2} - {\color{red}{\left(0\right)}}$$

Thus, $$$\frac{\partial}{\partial y}\left(- 4 x + 4 y^{3}\right)=12 y^{2}$$$

Therefore, $$$\frac{\partial^{2}}{\partial y^{2}}\left(x^{4} - 4 x y + y^{4} + 1\right)=12 y^{2}$$$

Answer: $$$\frac{\partial^{2}}{\partial y^{2}}\left(x^{4} - 4 x y + y^{4} + 1\right)=12 y^{2}$$$