Chain Rule
The Essentials
The formula for the chain rule for one independent variable is:
\[
\frac{dz}{dt}=\frac{\partial z}{\partial x}\cdot\frac{dx}{dt}+ \frac{\partial z}{\partial y}\cdot\frac{dy}{dt}
\]
For two independent variables:
\[
\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial u}
\]
\[
\frac{\partial z}{\partial v}=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial v}
\]
The generalized chain rule of a function \( w(x_1,x_2,\ldots,x_m) \) with any number of independent variables \( x_i=x_i(t_1,t_2,\ldots,t_n) \):
\[
\frac{\partial w}{\partial t_j}=\frac{\partial w}{\partial x_1}\cdot\frac{\partial x_1}{\partial t_j}+\frac{\partial w}{\partial x_2}\cdot\frac{\partial x_2}{\partial t_j}+\ldots+\frac{\partial w}{\partial x_m}\cdot\frac{\partial x_m}{\partial t_j}
\]
In a function of two variables \( f(x, y)=0 \) where \( y \) is an implicit function of \( x \), implicit differentiation can be used to find \( \frac{dy}{dx} \):
\[
\frac{dy}{dx}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}
\]
For a function of three variables \( f(x,y,z)=0 \) where \( z \) is implicitly a function of \( x \) and \( y \):
\[
\frac{\partial z}{\partial x}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}
\]
\[
\frac{\partial z}{\partial y}=-\frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial z}}
\]
Example
Find \( \frac{dy}{dx} \):
\[
x^2+xy+y^2-4=0
\]
First, we need to find the partial of the function with respect to \( x \), then with respect to \( y \):
\[
\frac{\partial f}{\partial x}= 2x+y
\]
\[
\frac{\partial f}{\partial y}= x+2y
\]
Now we can use the formula:
\[
\frac{dy}{dx}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}
\]
\[
\frac{dy}{dx}=-\frac{2x+y}{x+2y}
\]
Practice
1) Find the equation of the line tangent to the function at the point \( P(1,1) \)
\[
x^2-4xy+2y^2-2x+y+2=0
\]
2) Find \( \frac{\partial z}{\partial t} \) for the function \( z=f(x,y) \)
\[
\begin{align*}
z &= x^2y^3 + \frac{1}{x}y \\
x &= e^t \\
y &= t^3 \\
\end{align*}
\]
Solutions:
1) \( y=4x-3 \)
2)
\[
\begin{align*}
\frac{\partial z}{\partial x} &= 2xy^3 - \frac{1}{x^2}y & \frac{\partial x}{\partial t} &= e^t \\
\frac{\partial z}{\partial y} &= 3x^2y^2 + \frac{1}{x} & \frac{\partial y}{\partial t} &= 3t^2 \\
\end{align*}
\]
Let's first solve \( \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} \):
\[
\begin{align*}
\frac{\partial z}{\partial x}\frac{\partial x}{\partial t} &= (2xy^3-\frac{1}{x^2}y)(e^t) \\
&= (2e^tt^9-\frac{1}{e^{2t}}t^3)(e^t) \\
&= 2e^{2t}t^9-\frac{1}{e^t}t^3 \\
\end{align*}
\]
Next, let's solve \( \frac{\partial z}{\partial y}\frac{\partial y}{\partial t} \):
\[
\begin{align*}
\frac{\partial z}{\partial y}\frac{\partial y}{\partial t} &= (3x^2y^2+\frac{1}{x})(3t^2) \\
&= (3e^{2t}(t^3)^2+\frac{1}{e^t})(3t^2) \\
&= 9t^2e^{2t}(t^6)+\frac{3t^2}{e^t} \\
&= 9t^8e^{2t}+\frac{3t^2}{e^t} \\
\end{align*}
\]
Lastly, we add both together:
\[
\begin{align*}
\frac{\partial z}{\partial t} = 2e^{2t}t^9-\frac{1}{e^t}t^3+9t^8e^{2t}+\frac{3t^2}{e^t} \\
\end{align*}
\]
