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Kısmi Türev Hesaplayıcı
Kısmi türevleri adım adım hesaplayın
Bu çevrimiçi hesaplayıcı, adımları göstererek fonksiyonun kısmi türevini hesaplar. Herhangi bir integrasyon sırasını belirtebilirsiniz.
Solution
Your input: find $$$\frac{\partial^{2}}{\partial y^{2}}\left(x^{3} + 4 x y^{2} + 5 y^{3} - 10\right)$$$
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, $$$\frac{\partial^{2}}{\partial y^{2}}\left(x^{3} + 4 x y^{2} + 5 y^{3} - 10\right)=\frac{\partial}{\partial y} \left(\frac{\partial}{\partial y}\left(x^{3} + 4 x y^{2} + 5 y^{3} - 10\right) \right)=\frac{\partial}{\partial y}\left(y \left(8 x + 15 y\right)\right)$$$
Apply the product rule $$$\frac{\partial}{\partial y} \left(f \cdot g \right)=\frac{\partial}{\partial y} \left(f \right) \cdot g + f \cdot \frac{\partial}{\partial y} \left(g \right)$$$ with $$$f=y$$$ and $$$g=8 x + 15 y$$$:
$${\color{red}{\frac{\partial}{\partial y}\left(y \left(8 x + 15 y\right)\right)}}={\color{red}{\left(y \frac{\partial}{\partial y}\left(8 x + 15 y\right) + \frac{\partial}{\partial y}\left(y\right) \left(8 x + 15 y\right)\right)}}$$Apply the power rule $$$\frac{\partial}{\partial y} \left(y^{n} \right)=n\cdot y^{-1+n}$$$ with $$$n=1$$$, in other words $$$\frac{\partial}{\partial y} \left(y \right)=1$$$:
$$y \frac{\partial}{\partial y}\left(8 x + 15 y\right) + \left(8 x + 15 y\right) {\color{red}{\frac{\partial}{\partial y}\left(y\right)}}=y \frac{\partial}{\partial y}\left(8 x + 15 y\right) + \left(8 x + 15 y\right) {\color{red}{1}}$$The derivative of a sum/difference is the sum/difference of derivatives:
$$8 x + 15 y + y {\color{red}{\frac{\partial}{\partial y}\left(8 x + 15 y\right)}}=8 x + 15 y + y {\color{red}{\left(\frac{\partial}{\partial y}\left(8 x\right) + \frac{\partial}{\partial y}\left(15 y\right)\right)}}$$Apply the constant multiple rule $$$\frac{\partial}{\partial y} \left(c \cdot f \right)=c \cdot \frac{\partial}{\partial y} \left(f \right)$$$ with $$$c=15$$$ and $$$f=y$$$:
$$8 x + 15 y + y \left({\color{red}{\frac{\partial}{\partial y}\left(15 y\right)}} + \frac{\partial}{\partial y}\left(8 x\right)\right)=8 x + 15 y + y \left({\color{red}{\left(15 \frac{\partial}{\partial y}\left(y\right)\right)}} + \frac{\partial}{\partial y}\left(8 x\right)\right)$$Apply the power rule $$$\frac{\partial}{\partial y} \left(y^{n} \right)=n\cdot y^{-1+n}$$$ with $$$n=1$$$, in other words $$$\frac{\partial}{\partial y} \left(y \right)=1$$$:
$$8 x + 15 y + y \left(15 {\color{red}{\frac{\partial}{\partial y}\left(y\right)}} + \frac{\partial}{\partial y}\left(8 x\right)\right)=8 x + 15 y + y \left(15 {\color{red}{1}} + \frac{\partial}{\partial y}\left(8 x\right)\right)$$The derivative of a constant is 0:
$$8 x + 15 y + y \left(15 + {\color{red}{\frac{\partial}{\partial y}\left(8 x\right)}}\right)=8 x + 15 y + y \left(15 + {\color{red}{\left(0\right)}}\right)$$Thus, $$$\frac{\partial}{\partial y}\left(y \left(8 x + 15 y\right)\right)=8 x + 30 y$$$
Therefore, $$$\frac{\partial^{2}}{\partial y^{2}}\left(x^{3} + 4 x y^{2} + 5 y^{3} - 10\right)=8 x + 30 y$$$
Answer: $$$\frac{\partial^{2}}{\partial y^{2}}\left(x^{3} + 4 x y^{2} + 5 y^{3} - 10\right)=8 x + 30 y$$$