Calculatrice de dérivées partielles

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

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

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: $\frac{\partial^{2}}{\partial y^{2}}\left(x^{3} + 4 x y^{2} + 5 y^{3} - 10\right)=8 x + 30 y$