Applying The Chain Rule to Functions of Several Variables Examples 1

Applying The Chain Rule to Functions of Several Variables Examples 1

We saw some examples of applying the various chain rules for functions of several variables on the Applying The Chain Rule to Functions of Several Variables page. We will now look at some more examples. Once again, here are the generic chain rules for reference:

If $z = f(x, y)$ has continuous first partial derivatives and $x = x(t)$ and $y = y(t)$ are differentiable then:

(1)
\begin{align} \quad \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} \end{align}

If $z = f(x, y)$ is a two variable real-valued function with continuous first partial derivatives, and $x = x(s, t)$ and $y = y(s, t)$ are functions of $s$ and $t$ then:

(2)
\begin{align} \quad \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} \end{align}
(3)
\begin{align} \quad \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t} \end{align}

Example 1

Suppose that $w = f(x, y)$, $x = g(r, s)$, $y = h(r, t)$, $r = k(s, t)$, and $s = m(t)$. Compute $\frac{\partial w}{\partial t}$.

This problem has many layers with it. We start with the outermost layer by applying the Chain Rule Type 2:

(4)
\begin{align} \quad \frac{\partial w}{\partial t} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial t} \\ \quad \frac{\partial w}{\partial t} = f_1(x, y) \frac{\partial x}{\partial t} + f_2(x, y) \frac{\partial w}{\partial y} \end{align}

Now we must compute both $\frac{\partial x}{\partial t}$ and $\frac{\partial y}{\partial t}$. Applying the Chain Rule again and we have that:

(5)
\begin{align} \quad \frac{\partial x}{\partial t} = \frac{\partial x}{\partial r} \frac{\partial r}{\partial t} + \frac{\partial x}{\partial s} \frac{\partial s}{\partial t} \\ \quad \frac{\partial x}{\partial t} = g_1(r, s) \frac{\partial r}{\partial t} + g_2(r, s) \frac{\partial s}{\partial t} \\ \quad \frac{\partial x}{\partial t} = g_1(r, s) \left ( \frac{\partial r}{\partial s} \frac{\partial s}{\partial t} + \frac{\partial r}{\partial t} \frac{\partial t}{\partial t} \right ) + g_2(r, s) m'(t) \\ \quad \frac{\partial x}{\partial t} = g_1(r, s) \left ( k_1(s, t) m'(t) + k_2(s, t) \cdot 1\right ) + g_2(r, s) m'(t) \\ \quad \frac{\partial x}{\partial t} = g_1(r, s) \left ( k_1(s, t) m'(t) + k_2(s, t)\right ) + g_2(r, s) m'(t) \end{align}
(6)
\begin{align} \quad \frac{\partial y}{\partial t} = \frac{\partial y}{\partial r} \frac{\partial r}{\partial t} + \frac{\partial y}{\partial t} \frac{\partial t}{\partial t} \\ \quad \frac{\partial y}{\partial t} = h_1(r, t) \frac{\partial r}{\partial t} + h_2(r, t) \cdot 1 \\ \quad \frac{\partial y}{\partial t} = h_1(r, t) \left ( \frac{\partial r}{\partial s} \frac{\partial s}{\partial t} + \frac{\partial r}{\partial t} \frac{\partial t}{\partial t} \right ) + h_2(r, t) \\ \quad \frac{\partial y}{\partial t} = h_1(r, t) \left ( k_1(s, t) m'(t) + k_2(s, t) \cdot 1 \right ) + h_2(r, t) \\ \quad \frac{\partial y}{\partial t} = h_1(r, t) \left ( k_1(s, t) m'(t) + k_2(s, t) \right ) + h_2(r, t) \\ \end{align}

Putting this all together and we get that:

(7)
\begin{align} \quad \frac{\partial w}{\partial t} = f_1(x, y) \left [ g_1(r, s) \left ( k_1(s, t) m'(t) + k_2(s, t)\right ) + g_2(r, s) m'(t) \right ] + f_2(x, y) \left [ h_1(r, t) \left ( k_1(s, t) m'(t) + k_2(s, t) \right ) + h_2(r, t) \right ] \end{align}

Example 2

Compute the partial derivative $\frac{\partial}{\partial x} f(2x, 3y)$ and $\frac{\partial}{\partial x} f(2y, 3x)$.

Let's first compute $\frac{\partial}{\partial x} f(2x, 3y)$. If we let $u(x) = 2x$ and $v(y) = 3y$ then we have that $f(2x, 3y) = f(u(x), v(y))$. Applying the Chain Rule and we have that:

(8)
\begin{align} \quad \frac{\partial}{\partial x} f(2x, 3y) = \frac{\partial}{\partial x} f(u, v) = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} = 2f_1(u, v) + 0f_2(u, v) = 2f_1(u,v) = 2f_1(2x, 3y) \end{align}

Now let's compute $\frac{\partial}{\partial x} f(2y, 3x)$. If we let $w(y) = 2y$ and $z(x) = 3x$ then we have that $f(2y, 3x) = f(w(y), z(x))$. Applying the Chain Rule and we have that:

(9)
\begin{align} \quad \frac{\partial}{\partial x} f(2y, 3x) = \frac{\partial}{\partial x} f(w, z) = \frac{\partial f}{\partial w} \frac{\partial w}{\partial x} + \frac{\partial f}{\partial z} \frac{\partial z}{\partial x} =0 f_1(w, z) + 3f_2(w, z) = 3f_2(w, z) = 3f_2(2y, 3x) \end{align}