Chain Rule Type 1 for Functions of Several Variables Example

Chain Rule Type 1 for Functions of Several Variables Example

Recall from The Chain Rule Type 1 for Functions of Several Variables page that if $z = f(x, y)$ is a two variable real-valued function where $x = x(t)$, and $y = y(t)$ are differentiable functions in terms of the parameter $t$, 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}

More generally, if $z = f(x_1, x_2, ..., x_n)$ is an $n$ variable real-valued function where $x_1 = x_1(t)$, $x_2 = x_2(t)$, …, $x_n = x_n(t)$ are all differentiable functions in terms of the single parameter $t$, then:

(2)
\begin{align} \quad \frac{dz}{dt} = \frac{\partial z}{\partial x_1} \frac{d x_1}{dt} + \frac{\partial z}{\partial x_2} \frac{d x_2}{dt} + ... + \frac{\partial z}{\partial x_n} \frac{d x_n}{dt} \end{align}

We will now look at some examples of applying the chain rule type 1.

Example 1

Let $z = f(x,y) = 2xy + \cos (x + y)$, where $x = 3t^2$ and $y = e^{3t + 1}$. Find $\frac{dz}{dt}$.

We note that the partial derivatives of $f$ are $\frac{\partial z}{\partial x} = 2y - \sin (x + y)$ and $\frac{\partial z}{\partial y} = 2x - \sin (x + y)$. We also have that $\frac{dx}{dt} = 6t$ and $\frac{dy}{dt} = 3e^{3t+1}$. Therefore:

(3)
\begin{align} \quad \frac{dz}{dt} = \left ( 2y - \sin (x + y) \right ) \left ( 6t \right ) + \left ( 2x - \sin (x + y) \right ) \left ( 3e^{3t+1} \right ) \\ \quad \frac{dz}{dt} = \left ( 2e^{3t + 1} - \sin \left(3t^2 + e^{3t + 1} \right) \right ) \left ( 6t \right ) + \left ( 6t^2 - \sin \left (3t^2 + e^{3t + 1} \right) \right ) \left ( 3e^{3t+1} \right ) \\ \end{align}

Example 2

Let $w = f(x, y, z) = \log_2 (x^2 + 2y + xz^3 )$, where $x = (t + 1)^{10}$, $y = \sin t + 1$, and $z = e^{2t + 1}$. Find $\frac{dw}{dt}$.

We note that the partial derivatives of $f$ are $\frac{\partial w}{\partial x} = \frac{2x + z^3}{\ln (2) (x^2 + 2y + xz^3)}$, $\frac{\partial w}{\partial y} = \frac{2}{\ln (2) (x^2 + 2y + xz^3)}$, and $\frac{\partial w}{\partial z} = \frac{3xz^2}{\ln (2) (x^2 + 2y + xz^3)}$. We also have that $\frac{dx}{dt} = 10(t + 1)^9$, $\frac{dy}{dt} = \cos t$, and $\frac{dz}{dt} = 2e^{2t+1}$. Therefore:

(4)
\begin{align} \quad \quad \frac{dw}{dt} = \left ( \frac{2x + z^3}{\ln (2) (x^2 + 2y + xz^3)} \right ) \left ( 10(t + 1)^9 \right ) + \left ( \frac{2}{\ln (2) (x^2 + 2y + xz^3)} \right ) \left ( \cos t \right ) + \left ( \frac{3xz^2}{\ln (2) (x^2 + 2y + xz^3)} \right ) \left ( 2e^{2t+1} \right ) \\ \quad \quad \frac{dw}{dt} = \left ( \frac{2(t + 1)^{10} + e^{6t + 3}}{\ln (2) ((t + 1)^{20} + 2(\sin t + 1) + (t + 1)^{10}e^{6t + 3})} \right ) \left ( 10(t + 1)^9 \right )\\ + \left ( \frac{2}{\ln (2) ((t + 1)^{20} + 2(\sin t + 1) + (t + 1)^{10}e^{6t + 3})} \right ) \left ( \cos t \right )\\ + \left ( \frac{3(t + 1)^{10}e^{4t + 2}}{\ln (2) ((t + 1)^{20} + 2(\sin t + 1) + (t + 1)^{10}e^{6t + 3})} \right ) \left ( 2e^{2t+1} \right ) \end{align}