Partial Derivative Calculator

Your input: find $\frac{\partial^{2}}{\partial x \partial y}\left(e^{x y}\right)$

Write the function $e^{x y}$ as a composition of the two functions $u=g=x y$ and $f\left(u\right)=e^{u}$.

Apply the chain rule $\frac{\partial}{\partial x} \left(f\left(g\right) \right)=\frac{\partial}{\partial u} \left(f\left(u\right) \right) \cdot \frac{\partial}{\partial x} \left(g \right)$:

$${\color{red}{\frac{\partial}{\partial x}\left(e^{x y}\right)}}={\color{red}{\frac{\partial}{\partial u}\left(e^{u}\right) \frac{\partial}{\partial x}\left(x y\right)}}$$

The derivative of an exponential is $\frac{\partial}{\partial u} \left(e^{u} \right)=e^{u}$:

$${\color{red}{\frac{\partial}{\partial u}\left(e^{u}\right)}} \frac{\partial}{\partial x}\left(x y\right)={\color{red}{e^{u}}} \frac{\partial}{\partial x}\left(x y\right)$$

Return to the old variable:

$$e^{{\color{red}{u}}} \frac{\partial}{\partial x}\left(x y\right)=e^{{\color{red}{x y}}} \frac{\partial}{\partial x}\left(x y\right)$$

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 $f=x$:

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

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

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

Thus, $\frac{\partial}{\partial x}\left(e^{x y}\right)=y e^{x y}$

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 $g=e^{x y}$:

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

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

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

Write the function $e^{x y}$ as a composition of the two functions $u=g=x y$ and $f\left(u\right)=e^{u}$.

Apply the chain rule $\frac{\partial}{\partial y} \left(f\left(g\right) \right)=\frac{\partial}{\partial u} \left(f\left(u\right) \right) \cdot \frac{\partial}{\partial y} \left(g \right)$:

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

The derivative of an exponential is $\frac{\partial}{\partial u} \left(e^{u} \right)=e^{u}$:

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

Return to the old variable:

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

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

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

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

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

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

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

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