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  1. Beranda
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  3. Kalkulator: Kalkulus III
  4. Kalkulator Kalkulus

Kalkulator Turunan Parsial

Hitung turunan parsial langkah demi langkah

Kalkulator online ini akan menghitung turunan parsial dari fungsi, beserta langkah-langkahnya. Anda dapat menentukan urutan integrasi apa pun.

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 contact us.

Solution

Your input: find $$$\frac{\partial^{2}}{\partial y \partial x}\left(x^{3} + 4 x y^{2} + 5 y^{3} - 10\right)$$$

First, find $$$\frac{\partial}{\partial y}\left(x^{3} + 4 x y^{2} + 5 y^{3} - 10\right)$$$

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

$${\color{red}{\frac{\partial}{\partial y}\left(x^{3} + 4 x y^{2} + 5 y^{3} - 10\right)}}={\color{red}{\left(- \frac{\partial}{\partial y}\left(10\right) + \frac{\partial}{\partial y}\left(x^{3}\right) + \frac{\partial}{\partial y}\left(5 y^{3}\right) + \frac{\partial}{\partial y}\left(4 x y^{2}\right)\right)}}$$

Apply the constant multiple rule $$$\frac{\partial}{\partial y} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial y} \left(f \right)$$$ with $$$c=4 x$$$ and $$$f=y^{2}$$$:

$${\color{red}{\frac{\partial}{\partial y}\left(4 x y^{2}\right)}} - \frac{\partial}{\partial y}\left(10\right) + \frac{\partial}{\partial y}\left(x^{3}\right) + \frac{\partial}{\partial y}\left(5 y^{3}\right)={\color{red}{4 x \frac{\partial}{\partial y}\left(y^{2}\right)}} - \frac{\partial}{\partial y}\left(10\right) + \frac{\partial}{\partial y}\left(x^{3}\right) + \frac{\partial}{\partial y}\left(5 y^{3}\right)$$

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

$$4 x {\color{red}{\frac{\partial}{\partial y}\left(y^{2}\right)}} - \frac{\partial}{\partial y}\left(10\right) + \frac{\partial}{\partial y}\left(x^{3}\right) + \frac{\partial}{\partial y}\left(5 y^{3}\right)=4 x {\color{red}{\left(2 y^{-1 + 2}\right)}} - \frac{\partial}{\partial y}\left(10\right) + \frac{\partial}{\partial y}\left(x^{3}\right) + \frac{\partial}{\partial y}\left(5 y^{3}\right)=8 x y - \frac{\partial}{\partial y}\left(10\right) + \frac{\partial}{\partial y}\left(x^{3}\right) + \frac{\partial}{\partial y}\left(5 y^{3}\right)$$

The derivative of a constant is 0:

$$8 x y - {\color{red}{\frac{\partial}{\partial y}\left(10\right)}} + \frac{\partial}{\partial y}\left(x^{3}\right) + \frac{\partial}{\partial y}\left(5 y^{3}\right)=8 x y - {\color{red}{\left(0\right)}} + \frac{\partial}{\partial y}\left(x^{3}\right) + \frac{\partial}{\partial y}\left(5 y^{3}\right)$$

The derivative of a constant is 0:

$$8 x y + {\color{red}{\frac{\partial}{\partial y}\left(x^{3}\right)}} + \frac{\partial}{\partial y}\left(5 y^{3}\right)=8 x y + {\color{red}{\left(0\right)}} + \frac{\partial}{\partial y}\left(5 y^{3}\right)$$

Apply the constant multiple rule $$$\frac{\partial}{\partial y} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial y} \left(f \right)$$$ with $$$c=5$$$ and $$$f=y^{3}$$$:

$$8 x y + {\color{red}{\frac{\partial}{\partial y}\left(5 y^{3}\right)}}=8 x y + {\color{red}{\left(5 \frac{\partial}{\partial y}\left(y^{3}\right)\right)}}$$

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

$$8 x y + 5 {\color{red}{\frac{\partial}{\partial y}\left(y^{3}\right)}}=8 x y + 5 {\color{red}{\left(3 y^{-1 + 3}\right)}}=y \left(8 x + 15 y\right)$$

Thus, $$$\frac{\partial}{\partial y}\left(x^{3} + 4 x y^{2} + 5 y^{3} - 10\right)=y \left(8 x + 15 y\right)$$$

Next, $$$\frac{\partial^{2}}{\partial y \partial x}\left(x^{3} + 4 x y^{2} + 5 y^{3} - 10\right)=\frac{\partial}{\partial x} \left(\frac{\partial}{\partial y}\left(x^{3} + 4 x y^{2} + 5 y^{3} - 10\right) \right)=\frac{\partial}{\partial x}\left(y \left(8 x + 15 y\right)\right)$$$

Apply the constant multiple rule $$$\frac{\partial}{\partial x} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial x} \left(f \right)$$$ with $$$c=y$$$ and $$$f=8 x + 15 y$$$:

$${\color{red}{\frac{\partial}{\partial x}\left(y \left(8 x + 15 y\right)\right)}}={\color{red}{y \frac{\partial}{\partial x}\left(8 x + 15 y\right)}}$$

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

$$y {\color{red}{\frac{\partial}{\partial x}\left(8 x + 15 y\right)}}=y {\color{red}{\left(\frac{\partial}{\partial x}\left(8 x\right) + \frac{\partial}{\partial x}\left(15 y\right)\right)}}$$

The derivative of a constant is 0:

$$y \left({\color{red}{\frac{\partial}{\partial x}\left(15 y\right)}} + \frac{\partial}{\partial x}\left(8 x\right)\right)=y \left({\color{red}{\left(0\right)}} + \frac{\partial}{\partial x}\left(8 x\right)\right)$$

Apply the constant multiple rule $$$\frac{\partial}{\partial x} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial x} \left(f \right)$$$ with $$$c=8$$$ and $$$f=x$$$:

$$y {\color{red}{\frac{\partial}{\partial x}\left(8 x\right)}}=y {\color{red}{\left(8 \frac{\partial}{\partial x}\left(x\right)\right)}}$$

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

$$8 y {\color{red}{\frac{\partial}{\partial x}\left(x\right)}}=8 y {\color{red}{1}}$$

Thus, $$$\frac{\partial}{\partial x}\left(y \left(8 x + 15 y\right)\right)=8 y$$$

Therefore, $$$\frac{\partial^{2}}{\partial y \partial x}\left(x^{3} + 4 x y^{2} + 5 y^{3} - 10\right)=8 y$$$

Answer: $$$\frac{\partial^{2}}{\partial y \partial x}\left(x^{3} + 4 x y^{2} + 5 y^{3} - 10\right)=8 y$$$


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