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Kalkylator för partialderivator

Beräkna partiella derivator steg för steg

Den här onlinekalkylatorn beräknar funktionens partialderivata och visar stegen. Du kan ange vilken integrationsordning som helst.

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

First, rewrite the function: $$$\frac{\partial^{2}}{\partial x \partial y}\left(\left(x - y\right) \left(- x y + 1\right)\right)=\frac{\partial^{2}}{\partial x \partial y}\left(\left(- x + y\right) \left(x y - 1\right)\right)$$$

Expand the function: $$$\frac{\partial^{2}}{\partial x \partial y}\left(\left(- x + y\right) \left(x y - 1\right)\right)=\frac{\partial^{2}}{\partial x \partial y}\left(- x^{2} y + x y^{2} + x - y\right)$$$

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

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

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

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

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

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

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

The derivative of a constant is 0:

$$y^{2} + 1 - {\color{red}{\frac{\partial}{\partial x}\left(y\right)}} - \frac{\partial}{\partial x}\left(x^{2} y\right)=y^{2} + 1 - {\color{red}{\left(0\right)}} - \frac{\partial}{\partial x}\left(x^{2} y\right)$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The derivative of a constant is 0:

$$- 2 x + 2 y + {\color{red}{\frac{\partial}{\partial y}\left(1\right)}}=- 2 x + 2 y + {\color{red}{\left(0\right)}}$$

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

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

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


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