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Calculadora de Derivada Parcial

Calcular derivadas parciais passo a passo

Esta calculadora online calculará a derivada parcial da função, com as etapas mostradas. Você pode especificar qualquer ordem de integração.

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

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