Section 8.3 Implicit Differentiation
We can use our chain rules to produce another way looking at implicit differentiation. Assuming that the equation
\begin{equation*}
F(x,y) = 0
\end{equation*}
implicitly defines the function \(y=y(x)\text{,}\) recall that implicit differentiation gives us a way of finding a formula for \(\dfrac{dy}{dx}\text{.}\)
Example 8.3.1.
Use implicit differentiation to find a formula for \(\dfrac{dy}{dx}\) for the function \(y=y(x)\) implicitly defined by the equation
\begin{equation*}
x\cos(y) + y\cos(x) =x\text{.}
\end{equation*}
Answer.\(\dfrac{dy}{dx} = \dfrac{1-\cos(y)+y\sin(x)}{\cos(x)-x\sin(y)}\)
Differentiating both sides of the equation with respect to \(x\)
\begin{alignat*}{1}
\dfrac{d}{dx}\left(x\cos(y) + y\cos(x)\right) \amp = \dfrac{d}{dx}(x)\\
\left(-x\sin(y)\dfrac{dy}{dx} + \cos(y)\right) + \left(-y\sin(x)+\cos(x)\dfrac{dy}{dx}\right) \amp = 1\\
\dfrac{dy}{dx}\left(\cos(x)-x\sin(y)\right) \amp = 1-\cos(y) +y\sin(x)\\
\dfrac{dy}{dx} \amp = \dfrac{1-\cos(y)+y\sin(x)}{\cos(x)-x\sin(y)}
\end{alignat*}
To use the chain rules to find a formula for \(\dfrac{dy}{dx}\) for the function implicitly defined by the equation
\begin{equation}
F(x,y)=0\label{Eqn-Implicit_function}\tag{8.3.1}
\end{equation}
let
\begin{equation*}
z=F(x,y) \text{ and } x=x(t), \quad y=y(t).
\end{equation*}
Thus we can think of \(F\) as being a function of the one variable \(t\text{,}\) and so, by Chain Rule 1,
\begin{equation*}
\dfrac{dz}{dt} = \dfrac{\partial F}{\partial x}\dfrac{dx}{dt} + \dfrac{\partial F}{\partial y}\dfrac{dy}{dt}.
\end{equation*}
Now we are thinking of equation (8.3.1) as defining a function of one variable \(y=y(x)\text{,}\) so let \(x=x\) and \(y=y(x)\) and hence
\begin{equation}
\dfrac{dz}{dx} = \dfrac{\partial F}{\partial x}\times (1) + \dfrac{\partial F}{\partial y}\dfrac{dy}{dx} = F_x + F_y \dfrac{dy}{dx}.\tag{8.3.2}
\end{equation}
Returning to equation (8.3.1), on differentiating both sides with respect to \(x\text{,}\) we obtain
\begin{alignat*}{1}
\dfrac{dz}{dx} \amp= \dfrac{d}{dx}(0)\\
F_x + F_y \dfrac{dy}{dx} \amp = 0.
\end{alignat*}
from which we obtain, provided \(F_y\neq 0\text{,}\)
\begin{equation*}
\dfrac{dy}{dx} = -\dfrac{F_x}{F_y}.
\end{equation*}
Thus we can find a formula for \(\dfrac{dy}{dx}\) via partial differentiation as opposed to implicit differentiation. In summary:
Theorem 8.3.2.
If the equation \(F(x,y) = 0\) implicitly defines the function \(y=y(x)\) then
\begin{equation*}
\dfrac{dy}{dx} = -\dfrac{F_x}{F_y} \text{ provided } F_y \neq 0.\
\end{equation*}
Example 8.3.3.
Use partial differentiation to find a formula for \(\dfrac{dy}{dx}\) for the function \(y=y(x)\) implicitly defined by the equation
\begin{equation*}
x\cos(y) + y\cos(x) =x\text{.}
\end{equation*}
Answer.\(\dfrac{dy}{dx} = \dfrac{1-\cos(y)+y\sin(x)}{\cos(x)-x\sin(y)}\)
Let
\begin{equation*}
F(x,y) = x\cos(y) + y\cos(x) - x\text{.}
\end{equation*}
Then
\begin{equation*}
\dfrac{\partial F}{\partial x} = \cos(y)-y\sin(x)-1 \text{ and } \dfrac{\partial F}{\partial y} = -x\sin(y)+\cos(x).
\end{equation*}
Thus
\begin{alignat*}{1}
\dfrac{dy}{dx} \amp = -\dfrac{F_x}{F_y}\\
\amp = \dfrac{1-\cos(y)+y\sin(x)}{\cos(x)-x\sin(y)}.
\end{alignat*}
A similar argument can extend this result to functions of more than one variable. For example:
Theorem 8.3.4.
If \(F(x,y,z) = 0\) implicitly defines the function \(z=f(x,y)\) then
\begin{equation*}
\dfrac{\partial z}{\partial x} = -\dfrac{F_x}{F_z} \text{ and } \dfrac{\partial z}{\partial y} = -\dfrac{F_y}{F_z}, \text{ provided } F_z \neq 0. \
\end{equation*}
Example 8.3.5.
Use partial differentiation to find formulas for \(\dfrac{\partial z}{\partial s} \text{ and } \dfrac{\partial z}{\partial t}\) for the function \(z=z(s,t)\) implicitly defined by the equation
\begin{equation*}
z^2 +\cos(s) + \ln(st) = 3.
\end{equation*}
Answer.
\(\dfrac{\partial z}{\partial s} = \dfrac{\sin(s)-\frac{1}{s}}{2z}\)
\(\dfrac{\partial z}{\partial t} = -\dfrac{1}{2zt}\)
Let
\begin{equation*}
F(s,t,z) = z^2 + \cos(s) + \ln(st)-3\text{.}
\end{equation*}
Then
\begin{equation*}
F_s = -\sin(s)+\dfrac{1}{s}, \quad F_t=\dfrac{1}{t}, F_z=2z.
\end{equation*}
Thus
\begin{equation*}
\dfrac{\partial z}{\partial s} = -\dfrac{F_s}{F_z} = \dfrac{\sin(s)-\frac{1}{s}}{2z}
\end{equation*}
\begin{equation*}
\dfrac{\partial z}{\partial t} = -\dfrac{F_t}{F_z}=-\dfrac{1}{2zt}.
\end{equation*}
Exercises Example Tasks
1.
Using partial differentiation (as opposed to implicit differentiation) find \(\dfrac{\partial z}{\partial x}\) at \((x,y,z) = (1,2,1)\) when the function \(z(x,y)\) is defined by the equation
\begin{equation*}
(x-y)e^z + (y-z)e^x + (z-x)e^y = 0.
\end{equation*}