Implicit Differentiation

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Implicit Differentiation

Implicit and Explicit functions

An explicit function is an function expressed as \( y = f(x)\) such as

       \(y = 2x^3 + 5\)

\( y \) is defined implicitly if both \( x \) and \( y \) occur on the same side of the equation such as

        \( x^2 + y^2 = 4 \)

we can think of \( y \) as function of \( x \) and write:

        \( x^2 + y(x)^2 = 4 \)

Implicit differentiation

To find \( \frac{dy}{dx} \), we proceed as follows:

Take \( \frac{d}{dx} \) of both sides of the equation remembering to multiply by \( y' \) each time you see a \( y \) term.
Solve for \( y' \)

Example

Find \( \frac{dy}{dx} \) implicitly for the circle

\( x^2 + y^2 = 4 \)

Solution

\( \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}4 \)

or

\( 2x + 2yy' = 0 \)
Solving for \( y \), we get

        \( 2yy' = -2x \)

        \( y' = \frac{-2x}{2y} \)

        \( y' = \frac{-x}{y} \)

Example:

Find y\( y' \) at \( (2,2) \) if

\( xy + \frac{x}{y} = 5 \)

Solution:

\( (xy)' + (\frac{x}{y})' = (5)' \)

Using the product rule and the quotient rule we have

\( xy' + y + \frac{y - xy'}{y^2} = 0 \)
Now plugging in \( x = 2 \) and \( y = 2 \),

        \(2y' + 2 + \frac{2 - 2y'}{4} = 0 \)         Multiply both sides by 4

        \( 8y' + 8 + 2 - 2y' = 0 \)

        \( 6y' = -10 \)

        \( y' = \frac{-5}{3} \)

Exercises:

Let

\( 3x^2 - y^2 = 4x + y^2 \)

Find \( \frac{dy}{dx} \)
Find \( \frac{dy}{dx} \) at (-1,1) if

\( x + y = x^3 + y^3 \)
Find \( \frac{dy}{dx} \) if

\( x^2+ 3xy + y^2 = 1 \)
Find \( y'' \) if

\( x^2 - y^2 = 4 \)

Application

Example

Suppose that the demand function for a boat shop is given by

\( p = -0.01x^3 + x + 10,000 \)

Find the rate of change of \( x \) with respect to p when \( x = 20 \). A boat craftsman can think of this question as who fast will the number of boats she will need to build change as the price is increased. Solving for \( x \) in terms of \( p \) is nearly impossible. Instead, we can differentiate implicitly.

\( 1 = -0.03x^2 x' + x' \)

Now plug in 20 for \( x \) to get

\( 1 = -0.03(20)^2 x' + x' \)

\( = -12x' + x' = -11x' \)

\( x' = \frac{-1}{11} \)

The rate of change is \( \frac{-1}{11} \) boats per dollar increase.

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