Multivariate Chain Rule

Bivariate Chain Rule (1 variable dependency) §

Let $f(x,y)$ be a Bivariate Function, with $x=g(t)$ and $y=h(t)$. I.e. $f$ depends on x and y, which depend on $t$. So $f$ can be written as $f(g(t),h(t))$.

The chain rule is:

$$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$$

Bivariate Chain Rule (2 variable dependency) §

However, if $x$ and $y$ depend on the same two variables $s$ and $t$, then $f$ now depends on $t$ and $s$. So if $x=g(s,t)$ and $y=h(s,t)$, then $f$ can be rewritten as $f(g(s,t),h(s,t))$. Now the chain rule is, essentially, using a monovarietal chain rule, twice:

And since Bivariate Function only support partial derivates, we can only have partial derivates of $f$:

$$\frac{\partial f}{\partial s}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}$$

$$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$$

Examples §

1: 1-variable dependency Chain Rule for a Paraboloid

If $z=x^2-y^2$, $x=\sin (t)$,$y=\cos (t)$. Find $\frac{dz}{dt}$ at $t=\frac{\pi }{6}$

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$ $\frac{\partial z}{\partial x}=2x=2\sin (t)$ $\frac{\partial z}{\partial y}=-2y=-2\cos (t)$ $\frac{dx}{dt}=\cos (t)$ $\frac{dy}{dt}=-\sin (t)$ Hence, $\frac{dz}{dt}=2\sin (t)\cos (t)+-2\cos (t)\times -\sin (t)$ $\frac{dz}{dt}=4\sin (t)\cos (t)$ Now, to find it at $t=\frac{\pi }{6}$ $\frac{dz}{dt}=4\sin \left(\frac{\pi }{6}\right)\cos \left(\frac{\pi }{6}\right)$ $=4\times \frac{1}{2}\times \frac{\sqrt{3}}{2}=\sqrt{3}$

2: 2-variable dependancy Chain Rule

If $z=e^x\sinh (y)$, $x=s{t}^2$, $y=s^{2}t$. Find $\frac{\partial z}{\partial s}$

$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$ $\frac{\partial z}{\partial x}=e^x\sinh (y)=e^{s{t}^2}\sinh (s^{2}t)$ $\frac{\partial z}{\partial y}=e^x\cosh (y)=e^{s{t}^2}\cosh (s^{2}t)$ $\frac{\partial x}{\partial s}=t^2$ $\frac{\partial y}{\partial s}=2st$ Hence, $\frac{\partial z}{\partial s}=e^{s{t}^2}(\sinh (s^{2}t)t^2+2st\cosh (s^{2}t))$