Implicit Differentiation

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Section 8.3 Implicit Differentiation

We can use our chain rules to produce another way looking at implicit differentiation. Assuming that the equation

F (x, y) = 0

implicitly defines the function

y = y (x),

recall that implicit differentiation gives us a way of finding a formula for

\frac{d y}{d x} .

Example 8.3.1.

Use implicit differentiation to find a formula for $\frac{d y}{d x}$ for the function $y = y (x)$ implicitly defined by the equation

x \cos (y) + y \cos (x) = x .

$\frac{d y}{d x} = \frac{1 - \cos (y) + y \sin (x)}{\cos (x) - x \sin (y)}$

Differentiating both sides of the equation with respect to $x$

\begin{aligned} \frac{d}{d x} (x \cos (y) + y \cos (x)) & = \frac{d}{d x} (x) \\ (- x \sin (y) \frac{d y}{d x} + \cos (y)) + (- y \sin (x) + \cos (x) \frac{d y}{d x}) & = 1 \\ \frac{d y}{d x} (\cos (x) - x \sin (y)) & = 1 - \cos (y) + y \sin (x) \\ \frac{d y}{d x} & = \frac{1 - \cos (y) + y \sin (x)}{\cos (x) - x \sin (y)} \end{aligned}

To use the chain rules to find a formula for

\frac{d y}{d x}

for the function implicitly defined by the equation

\begin{matrix} (8.3.1) & F (x, y) = 0 \end{matrix}

let

z = F (x, y) and x = x (t), y = y (t) .

Thus we can think of

F

as being a function of the one variable

t,

and so, by Chain Rule 1,

\frac{d z}{d t} = \frac{\partial F}{\partial x} \frac{d x}{d t} + \frac{\partial F}{\partial y} \frac{d y}{d t} .

Now we are thinking of equation (8.3.1) as defining a function of one variable

y = y (x),

so let

x = x

and

y = y (x)

and hence

\begin{matrix} (8.3.2) & \frac{d z}{d x} = \frac{\partial F}{\partial x} \times (1) + \frac{\partial F}{\partial y} \frac{d y}{d x} = F_{x} + F_{y} \frac{d y}{d x} . \end{matrix}

Returning to equation (8.3.1), on differentiating both sides with respect to

x,

we obtain

\begin{aligned} \frac{d z}{d x} & = \frac{d}{d x} (0) \\ F_{x} + F_{y} \frac{d y}{d x} & = 0. \end{aligned}

from which we obtain, provided

F_{y} \neq 0,

\frac{d y}{d x} = - \frac{F_{x}}{F_{y}} .

Thus we can find a formula for

\frac{d y}{d x}

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

\frac{d y}{d x} = - \frac{F_{x}}{F_{y}} provided F_{y} \neq 0.

Example 8.3.3.

Use partial differentiation to find a formula for $\frac{d y}{d x}$ for the function $y = y (x)$ implicitly defined by the equation

x \cos (y) + y \cos (x) = x .

$\frac{d y}{d x} = \frac{1 - \cos (y) + y \sin (x)}{\cos (x) - x \sin (y)}$

Let

F (x, y) = x \cos (y) + y \cos (x) - x .

Then

\frac{\partial F}{\partial x} = \cos (y) - y \sin (x) - 1 and \frac{\partial F}{\partial y} = - x \sin (y) + \cos (x) .

Thus

\begin{aligned} \frac{d y}{d x} & = - \frac{F_{x}}{F_{y}} \\ = \frac{1 - \cos (y) + y \sin (x)}{\cos (x) - x \sin (y)} . \end{aligned}

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

\frac{\partial z}{\partial x} = - \frac{F_{x}}{F_{z}} and \frac{\partial z}{\partial y} = - \frac{F_{y}}{F_{z}}, provided F_{z} \neq 0.

Example 8.3.5.

Use partial differentiation to find formulas for $\frac{\partial z}{\partial s} and \frac{\partial z}{\partial t}$ for the function $z = z (s, t)$ implicitly defined by the equation

z^{2} + \cos (s) + \ln (s t) = 3.

$\frac{\partial z}{\partial s} = \frac{\sin (s) - \frac{1}{s}}{2 z}$

$\frac{\partial z}{\partial t} = - \frac{1}{2 z t}$

Let

F (s, t, z) = z^{2} + \cos (s) + \ln (s t) - 3 .

Then

F_{s} = - \sin (s) + \frac{1}{s}, F_{t} = \frac{1}{t}, F_{z} = 2 z .

Thus

\frac{\partial z}{\partial s} = - \frac{F_{s}}{F_{z}} = \frac{\sin (s) - \frac{1}{s}}{2 z}

\frac{\partial z}{\partial t} = - \frac{F_{t}}{F_{z}} = - \frac{1}{2 z t} .

Exercises Example Tasks

1.

Using partial differentiation (as opposed to implicit differentiation) find

\frac{\partial z}{\partial x}

at

(x, y, z) = (1, 2, 1)

when the function

z (x, y)

is defined by the equation

(x - y) e^{z} + (y - z) e^{x} + (z - x) e^{y} = 0.