全微分可能な関数 $z = f(x,y)$ に対し, $x$ と $y$ がともに $t$ の関数で $t$ で微分可能である時

$\dfrac{dz}{dt} = \dfrac{\partial z}{\partial x} \dfrac{dx}{dt} + \dfrac{\partial z}{\partial y}\dfrac{dy}{dt}$

が成り立ちます。

$z = x^2+y^3$ より

$\dfrac{\partial z}{\partial x} = 2x,~~\dfrac{\partial z}{\partial y} = 3y^2$

また

$\dfrac{dx}{dt} = e^t,~~\dfrac{dy}{dt} = -e^{-t}$

よって

$\dfrac{dz}{dt} = \dfrac{\partial z}{\partial x} \dfrac{dx}{dt} + \dfrac{\partial z}{\partial y}\dfrac{dy}{dt} = 2e^t \cdot e^t + 3e^{-2t}(-e^{-t}) = 2e^{2t} - 3e^{-3t}$

となります。

$z = f(x,y)$ に対し, $x$ と $y$ がともに $u$, $v$ に関する $2$ 変数関数で, かつ偏微分可能である時

$\dfrac{\partial z}{\partial u} = \dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial u} + \dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial u}$

$\dfrac{\partial z}{\partial v} = \dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v} + \dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v}$

が成り立ちます。

$z = \dfrac{1}{x+y}$ より

$\dfrac{\partial z}{\partial x} = -\dfrac{1}{(x+y)^2},~~\dfrac{\partial z}{\partial y} = -\dfrac{1}{(x+y)^2}$

また

$\dfrac{\partial x}{\partial u} = v,~~\dfrac{\partial x}{\partial v} = u,~~\dfrac{\partial y}{\partial u} = 1,~~\dfrac{\partial y}{\partial v}=1$

であるから

$\begin{eqnarray*}\dfrac{\partial z}{\partial u} & = & \dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial u} + \dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial u}\\[1em] & = & -\dfrac{v}{(uv+u+v)^2} - \dfrac{1}{(uv+u+v)^2}\\[1em] & = & -\dfrac{v+1}{(uv+u+v)^2} \end{eqnarray*}$

同様に

$\begin{eqnarray*}\dfrac{\partial z}{\partial v} & = & \dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v} + \dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v}\\[1em] & = & -\dfrac{u}{(uv+u+v)^2} - \dfrac{1}{(uv+u+v)^2}\\[1em] & = & -\dfrac{u+1}{(uv+u+v)^2} \end{eqnarray*}$

となります。

$x = a + ht,~~y = b + kt$ であるから

$\dfrac{dx}{dt} = h,~~\dfrac{dy}{dt} = k$

よって

$\cfrac{df}{dt}=\cfrac{\partial f}{\partial x}\cfrac{dx}{dt}+\cfrac{\partial f}{\partial y} \cfrac{dy}{dt} = hf_x(a + ht,\ b + kt) + kf_y(a + ht,\ b + kt)$

$x=r\cos \theta$ であるから

$\dfrac{\partial x}{\partial r} = \cos \theta,~~\dfrac{\partial x}{\partial \theta} = -r\sin \theta$

また, $y = r\sin \theta$ であるから

$\dfrac{\partial y}{\partial r} = \sin \theta,~~\dfrac{\partial y}{\partial \theta} = r\cos \theta$

よって

$\dfrac{\partial f}{\partial r} = \dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial r} + \dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial r} = f_x\cos \theta + f_y\sin \theta$

また

$\dfrac{\partial f}{\partial \theta} = \dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial \theta} + \dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial \theta} = -f_xr\sin \theta + f_yr\cos \theta$

となります。