The Multivariable Chain Rule

$w \to x, y \to t$; no direct $t$.

$\dfrac{dw}{dt} = w_x\,\dfrac{dx}{dt} + w_y\,\dfrac{dy}{dt}$.

$w_x = 2xy$, $\ w_y = x^2 - 3y^2$, $\ \dfrac{dx}{dt} = -\sin t$, $\ \dfrac{dy}{dt} = \cos t$.

At $t = \tfrac{\pi}{4}$, $x = y = \tfrac{\sqrt{2}}{2}$: $w_x = 1$, $w_y = \tfrac{1}{2} - \tfrac{3}{2} = -1$. So $\dfrac{dw}{dt} = 1\left(-\tfrac{\sqrt{2}}{2}\right) + (-1)\tfrac{\sqrt{2}}{2} = -\sqrt{2}$.

$w \to x, y \to t$.

$\dfrac{dw}{dt} = f_x\,\dfrac{dx}{dt} + f_y\,\dfrac{dy}{dt}$.

$\dfrac{dx}{dt} = 2t = 2$, $\ \dfrac{dy}{dt} = 3$ at $t = 1$.

At $t = 1$ the point is $(x, y) = (2, 3)$, exactly where the data is given: $4 \cdot 2 + (-1) \cdot 3 = 5$.

$w \to x \to t$, $\ w \to y \to t$, and a direct branch $w \to t$.

$\dfrac{dw}{dt} = w_x\,\dfrac{dx}{dt} + w_y\,\dfrac{dy}{dt} + w_t$.

$w_x = 2x$, $\ w_y = t$, $\ w_t = y$, $\ \dfrac{dx}{dt} = \cos t$, $\ \dfrac{dy}{dt} = e^{t}$.

$\dfrac{dw}{dt} = 2\sin t\cos t + te^{t} + e^{t}$; at $t = 0$: $0 + 0 + 1 = 1$.

$w \to x, y$; each of $x, y \to s, t$.

$w_s = w_x x_s + w_y y_s$, $\ w_t = w_x x_t + w_y y_t$.

$x_s = t$, $x_t = s$, $y_s = 1$, $y_t = -1$. At $(1, 2)$: $x = 2$, $y = -1$, so $w_x = 2x + y = 3$, $w_y = x = 2$.

$w_s = 3 \cdot 2 + 2 \cdot 1 = 8$, $\ w_t = 3 \cdot 1 + 2 \cdot (-1) = 1$.

$z \to g, h$; each of $g, h \to s, t$.

$f_1$ and $f_2$ are the partials of $f$ in its first and second slots: $z_s = f_1\,g_s + f_2\,h_s$, $\ z_t = f_1\,g_t + f_2\,h_t$.

Inner point: $(g, h) = (1, 4)$, so $f_1 = 5$, $f_2 = -2$. Inner partials: $g_s = 3$, $g_t = -1$, $h_s = 0$, $h_t = 2$.

$z_s = 5 \cdot 3 + (-2) \cdot 0 = 15$, $\ z_t = 5 \cdot (-1) + (-2) \cdot 2 = -9$.

$g \to$ (first slot $u^2 - v$, second slot $uv$) $\to u, v$.

$g_u = f_1 \cdot \dfrac{\partial}{\partial u}(u^2 - v) + f_2 \cdot \dfrac{\partial}{\partial u}(uv)$, and the same with $v$.

$\dfrac{\partial}{\partial u}$: $2u$ and $v$. $\ \dfrac{\partial}{\partial v}$: $-1$ and $u$.

$g_u = 2u\,f_1 + v\,f_2$, $\ g_v = -f_1 + u\,f_2$, with $f_1, f_2$ at $(u^2 - v, uv)$.

$u = x - 3y$, so $z = F(u)$.

$z_x = F'(u) \cdot 1$, $\ z_y = F'(u) \cdot (-3)$.

$3z_x + z_y = 3F'(u) - 3F'(u)$.

$= 0$ for every $F$. ✓

$u = \dfrac{x}{y}$, $\ v = \dfrac{y}{z}$.

$w_x = \dfrac{1}{y}f_u$, $\ w_y = -\dfrac{x}{y^2}f_u + \dfrac{1}{z}f_v$, $\ w_z = -\dfrac{y}{z^2}f_v$.

$xw_x + yw_y + zw_z = \dfrac{x}{y}f_u - \dfrac{x}{y}f_u + \dfrac{y}{z}f_v - \dfrac{y}{z}f_v$.

Everything cancels: $0$. ✓

$s = \dfrac{y}{x}$, so $u = x^2F(s)$.

Product rule: $u_x = 2xF(s) + x^2F'(s)\left(-\dfrac{y}{x^2}\right) = 2xF(s) - yF'(s)$; $\ u_y = x^2F'(s) \cdot \dfrac{1}{x} = xF'(s)$.

$xu_x + yu_y = 2x^2F(s) - xyF'(s) + xyF'(s)$.

$= 2x^2F(s) = 2u$. ✓

$u = x + 2y$, $\ v = 2x - y$.

$w_x = f_u + 2f_v$, $\ w_y = 2f_u - f_v$.

$(f_u + 2f_v)^2 + (2f_u - f_v)^2 = 5f_u^2 + 5f_v^2$ (the $\pm 4f_uf_v$ cancel).

Match: $\alpha = 5$.

$g_s = f_1 + 3f_2$.

The inner derivatives $1$ and $3$ are constants, so no product-rule terms: $g_{st} = \dfrac{\partial}{\partial t}f_1 + 3\,\dfrac{\partial}{\partial t}f_2$.

$\dfrac{\partial}{\partial t}f_1 = 2f_{11} - f_{12}$, $\ \dfrac{\partial}{\partial t}f_2 = 2f_{21} - f_{22}$.

$g_{st} = 2f_{11} - f_{12} + 6f_{21} - 3f_{22} = 2f_{11} + 5f_{12} - 3f_{22}$.

$\dfrac{dz}{dt} = 2t\,f_x + 3\,f_y$.

$\dfrac{d^2z}{dt^2} = 2f_x + 2t\,\dfrac{d}{dt}f_x + 3\,\dfrac{d}{dt}f_y$.

$\dfrac{d}{dt}f_x = 2t\,f_{xx} + 3f_{xy}$, $\ \dfrac{d}{dt}f_y = 2t\,f_{yx} + 3f_{yy}$.

At $t = 1$, $(x, y) = (1, 3)$: $2(2) + 2(4 \cdot 2 + 1 \cdot 3) + 3(1 \cdot 2 + 3 \cdot 3) = 4 + 22 + 33 = 59$.

$u_t = -3f'(x - 3t) + 3g'(x + 3t)$, $\ u_x = f'(x - 3t) + g'(x + 3t)$.

The inner derivatives $\pm 3$ and $1$ are constants: no product-rule terms.

$u_{tt} = 9f''(x - 3t) + 9g''(x + 3t)$, $\ u_{xx} = f''(x - 3t) + g''(x + 3t)$.

$u_{tt} = 9\big(f'' + g''\big) = 9u_{xx}$. ✓