Interesting Differentiation Technique

Look at $y = f(u(x),v(x))$: $$\frac{dy}{dx} = \frac{\partial f}{\partial u}\cdot\frac{du}{dx}+\frac{\partial f}{\partial v}\cdot\frac{dv}{dx}$$ Now, note that $x^{\sin x}$ can be written as: $$y=x^{\sin x} = f(x,\sin x), \ f(u,v) = u^v$$ So that: $$\frac{dy}{dx} = \frac{\partial f}{\partial u}\cdot\frac{du}{dx}+\frac{\partial f}{\partial v}\cdot\frac{dv}{dx}= vu^{v-1} \cdot\frac{du}{dx} + \log u \cdot u^v \cdot\frac{dv}{dx}$$ In fact, this holds for any number of terms (since so does the chain rule), where each time we only look at the derivative by $x$ of only one of the terms, multiply those by the "internal" derivative, and sum them all up.

I relied on something similar to this in a published paper. There's an identity that, in the concrete instance where the number of independent variables is $3$, says \begin{align} & \phantom{{}=} \frac{\partial^3}{\partial x_1\,\partial x_2\,\partial x_3} e^y \\[10pt] & = e^y\left(\frac{\partial^3 y}{\partial x_1\,\partial x_2\,\partial x_3} + \frac{\partial^2 y}{\partial x_1\,\partial x_2}\cdot\frac{\partial y}{\partial x_3} + \frac{\partial^2 y}{\partial x_1\,\partial x_3}\cdot\frac{\partial y}{\partial x_2} + {}\right. \\[10pt] & \left.\phantom{{}= e^y\quad{}} + \frac{\partial^2 y}{\partial x_2\,\partial x_3}\cdot\frac{\partial y}{\partial x_1} + \frac{\partial y}{\partial x_1}\cdot\frac{\partial y}{\partial x_2}\cdot\frac{\partial y}{\partial x_3} \right) \end{align}

The point is there's one term for each partition of the set of variables. Having proved this, one can go on to say that $$ \frac{d^3}{dx^3} e^y = e^y\left( \frac{d^3 y}{dx^3} + 3\frac{d^2y}{dx^2}\cdot\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 \right), $$ simply by saying that's the special case in which all three variables are the same. The proof is the same, but it's clearer when one first treats the variables as distinguishable. When it's written in that form, one can see that there's just one term for each set partition, and all the coefficients are $1$, so that the coefficients in the form with indistinguishable terms have a combinatorial interpreation as the number of set partitions corresponding to a given integer partition.

Similarly \begin{align} & {}\qquad\frac{\partial^3}{\partial x_1\,\partial x_2\,\partial x_3} (uv) \\[10pt] & = u\frac{\partial^3 v}{\partial x_1\,\partial x_2\,\partial x_3} + \frac{\partial u}{\partial x_1}\cdot\frac{\partial^2 v}{\partial x_2\,\partial x_3} + \frac{\partial u}{\partial x_2}\cdot\frac{\partial^2 v}{\partial x_1\,\partial x_3} + \frac{\partial u}{\partial x_3}\cdot\frac{\partial^2 v}{\partial x_1\,\partial x_2} \\[10pt] & \phantom{{}=} + \frac{\partial^2 u}{\partial x_1\,\partial x_2}\cdot\frac{\partial v}{\partial x_3} + \frac{\partial^2 u}{\partial x_1\,\partial x_3}\cdot\frac{\partial v}{\partial x_3} + \frac{\partial^2 u}{\partial x_2\,\partial x_3}\cdot\frac{\partial v}{\partial x_1} + \frac{\partial^3 u}{\partial x_1\,\partial x_2\,\partial x_3}\cdot v \end{align} This time, there is one term for each subset of the set of variables. Each coefficient is $1$. Then one can make all variables indistinguishable, and collect like terms, and then each coefficient is the number of subsets of a specified size.