- #1
- 996
- 5
Not homework, just having fun. Every reference I find illustrates the chain rule for composite functions of two variables in this way:
[tex]
\begin{align*}
B &= f(x,y) \\
x &= g(w,z) \\
y &= h(w,z) \\
\frac{\partial B}{\partial w} &= \left( \frac{\partial B}{\partial x} \cdot \frac{\partial x}{\partial w} \right ) + \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial w}\right) \\
\frac{\partial B}{\partial z} &= \left(\frac{\partial B}{\partial x} \cdot \frac{\partial x}{\partial z}\right) + \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial z}\right)
\end{align*}
[/tex]
But, my situation is:
[tex]
\begin{align*}
B &= f(x,y) \\
y &= g(x,z)
\end{align*}
[/tex]
Is the following correct (since x is an independent variable)?
[tex]
\begin{align*}
\frac{\partial B}{\partial x} &= \left(\frac{\partial B}{\partial x} \cdot \frac{\partial x}{\partial x}\right) + \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial x}\right) \\
&= \left(\frac{\partial B}{\partial x} \cdot 1\right) + \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial x}\right) \\
&= \left(\frac{\partial B}{\partial x}\right) + \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial x}\right) \\ \\
\frac{\partial B}{\partial z} &= \left(\frac{\partial B}{\partial x} \cdot \frac{\partial x}{\partial z}\right) + \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial z}\right) \\
&= \left(\frac{\partial B}{\partial x} \cdot 0\right) + \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial z}\right) \\
&= \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial z}\right)
\end{align*}
[/tex]
Looks kind of goofy, but seems to work using the following example:
[tex]
\begin{align*}
B &= x^2y + xy \\
y &= x^3z
\end{align*}
[/tex]
[tex]
\begin{align*}
B &= f(x,y) \\
x &= g(w,z) \\
y &= h(w,z) \\
\frac{\partial B}{\partial w} &= \left( \frac{\partial B}{\partial x} \cdot \frac{\partial x}{\partial w} \right ) + \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial w}\right) \\
\frac{\partial B}{\partial z} &= \left(\frac{\partial B}{\partial x} \cdot \frac{\partial x}{\partial z}\right) + \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial z}\right)
\end{align*}
[/tex]
But, my situation is:
[tex]
\begin{align*}
B &= f(x,y) \\
y &= g(x,z)
\end{align*}
[/tex]
Is the following correct (since x is an independent variable)?
[tex]
\begin{align*}
\frac{\partial B}{\partial x} &= \left(\frac{\partial B}{\partial x} \cdot \frac{\partial x}{\partial x}\right) + \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial x}\right) \\
&= \left(\frac{\partial B}{\partial x} \cdot 1\right) + \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial x}\right) \\
&= \left(\frac{\partial B}{\partial x}\right) + \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial x}\right) \\ \\
\frac{\partial B}{\partial z} &= \left(\frac{\partial B}{\partial x} \cdot \frac{\partial x}{\partial z}\right) + \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial z}\right) \\
&= \left(\frac{\partial B}{\partial x} \cdot 0\right) + \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial z}\right) \\
&= \left(\frac{\partial B}{\partial y} \cdot \frac{\partial y}{\partial z}\right)
\end{align*}
[/tex]
Looks kind of goofy, but seems to work using the following example:
[tex]
\begin{align*}
B &= x^2y + xy \\
y &= x^3z
\end{align*}
[/tex]
Last edited: