- Given the following functions and their dependencies:
$$ u(r,v,w) \\ v(r,t) \\ w(r,t) $$
What is the chain rule expansion of $$\frac{\partial u}{\partial r}$$
- Given the following functions and their dependencies:
$$ y(x,\mu,\sigma) \\ \mu(x) \\ \sigma(x,\mu) $$
What is the chain rule expansion of $$\frac{\partial y}{\partial x}$$
I am stuck at writing down the first term itself. If I write $\frac{\partial u}{\partial r} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial r} + \dots$, then I get $\frac{\partial u}{\partial r}$ on both sides of the equation, which is incorrect.
\begin{equation} u=u(r,v(r,t),w(r,t)) \\ \end{equation} Because $v$ and $w$ depend on $r$ you can't just take $u$'s partial derivative . Instead you need to take its total derivative w.r.t $r$:
$$\dfrac{\mathrm{du}}{\mathrm{dr}}=\frac{\partial u}{\partial r}\underbrace{\frac{\mathrm{dr}}{\mathrm{dr}}}_{1}+\frac{\partial u}{\partial v}\frac{\mathrm{dv}}{\mathrm{dr}}+\frac{\partial u}{\partial w}\frac{\mathrm{dw}}{\mathrm{dr}}$$
$$\dfrac{\mathrm{du}}{\mathrm{dr}}=\frac{\partial u}{\partial r}+\frac{\partial u}{\partial v}\frac{\mathrm{dv}}{\mathrm{dr}}+\frac{\partial u}{\partial w}\frac{\mathrm{dw}}{\mathrm{dr}}$$
For $t$ it's a bit different since $u$ does not depend on $t$ directly:
$$\dfrac{\mathrm{du}}{\mathrm{dt}}=\frac{\partial u}{\partial r}\underbrace{{\frac{\mathrm{dr}}{\mathrm{dt}}}}_{0}+\frac{\partial u}{\partial v}{\frac{\mathrm{du}}{\mathrm{dt}}}+\frac{\partial u}{\partial w}{\frac{\mathrm{dw}}{\mathrm{dt}}}$$
$$\dfrac{\mathrm{du}}{\mathrm{dt}}=\frac{\partial u}{\partial v}{\frac{\mathrm{dv}}{\mathrm{dt}}}+\frac{\partial u}{\partial w}{\frac{\mathrm{dw}}{\mathrm{dt}}}$$