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Implicit Differentiation

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    217636
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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:

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

    2. Solve for \( y' \)

    Example

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

            \( x^2 + y^2 = 4 \)

    Solution

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

      or

              \( 2x + 2yy'  =  0 \)

    2. 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:  
     

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

      Using the product rule and the quotient rule we have

      \( xy' + y + \frac{y - xy'}{y^2}   =  0 \)

    2. 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:

    1. Let    

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

      Find \( \frac{dy}{dx} \)

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

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

    3. Find \( \frac{dy}{dx} \) if

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

    4. 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.

     


    Back to the Differentiation home page

     

     

    Implicit Differentiation is shared under a CC BY license and was authored, remixed, and/or curated by LibreTexts.

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