Derivative of bivariate function

Question

Let $G: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ be differentiable. According to mathmatica it holds $$ \frac{d}{d x} G(x,x) = G^{(0,1)}(x,x) + G^{(1,0)}(x,x) , $$ where I suppose that $G^{(0,1)}(x,y) = \frac{d}{d y} G(x,y)$ . How can this equality be derived?

Answer 1

There are two possible ways.

One way is to apply the chain rule. The map $F: x \mapsto (x, x)$ is differentiable with $F' = (1, 1)^T$. $G$ is differentiable with derivative $G'(x, y) = (G^{(1, 0)}(x, y), G^{(0, 1)}(x, y))$. Now the derivative of $G \circ F$ is given by

$$\begin{align*} (G \circ F)'(x) &= G'(F(x)) \cdot F'(x) = (G^{(1, 0)}(x, x), G^{(0, 1)}(x, x))\cdot (1, 1)^T \\ &= G^{(1, 0)}(x, x) + G^{(0, 1)}(x, x) \end{align*}$$

Alternatively, you can interpret $\frac{d}{dx} G(x, x)$ as the directional derivative of $G$ at (x, x) in the direction $(1, 1)^T$, since $$\frac{d}{dx} G(x, x) = \lim \limits_{h \to 0} \frac{G(x + h, x + h) - G(x, x)}{h}.$$ By the standard formula for directional derivatives we have $$\frac{d}{dx} G(x, x) = \nabla G(x, x) \cdot (1, 1)^T = G^{(1, 0)}(x, x) + G^{(0, 1)}(x, x).$$