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Implicit Differentiation

PROBLEM SOLVING STRATEGY: Implicit Differentiation

There are two basic steps to solve implicit differentiation problems:

  1. Take the derivative  $\dfrac{d}{dx}$  of both sides of the equation.
    • Use your usual Rules of Differentiation, with one addition: When you take the derivative of a term with a y in it, be sure to multiply by  $\dfrac{dy}{dx}$  due to the Chain Rule.
      Why do we multiply by dy/dx? Open for an explanation.
      Let’s consider, as an example, the function   $e^{x^2}$.   When you take its derivative, you of course use the Chain Rule:

      \dfrac{d}{dx}\left(e^{x^2} \right) &= e^{x^2} \cdot \dfrac{d}{dx}\left(x^2 \right) \\[8px]&= e^{x^2} \cdot (2x)


      Similarly, when you take the derivative of, say,   $e^{f(x)}$,  where $f(x)$ is some function we’re not specifying, you again of course use the Chain Rule:

      \dfrac{d}{dx}\left(e^{f(x)} \right) &= e^{f(x)} \cdot \dfrac{d}{dx}\left(f(x) \right) \\[8px]&= e^{f(x)} \cdot \dfrac{df}{dx}
      That might look funny, but we don’t know what  $\dfrac{df}{dx}$  is, so we’re just leaving it written like that.

      And also similarly, let’s consider the derivative of, say,  $e^{y(x)}$,  where we’re writing $y(x)$ to indicate that y is a function of x. We again of course must use the Chain Rule:

      \dfrac{d}{dx}\left(e^{y(x)} \right) &= e^{y(x)} \cdot \dfrac{d}{dx}\left(y(x) \right) \\[8px]&= e^{y(x)} \cdot \dfrac{dy(x)}{dx}
      Now we don’t usually write $y(x)$, because it’s annoying to write again and again; instead we just write “y”—but always have to keep in mind that really y still depends on x. We thus usually write that last result as

      $$\dfrac{d}{dx}\left(e^y \right) = e^y \cdot \dfrac{dy}{dx}$$

      The key point is that we end up multiplying by  $\dfrac{dy}{dx}$,  because the function we were taking the derivative of,  $e^y$,  has a y in it. Similarly, you should see that the following are all correct derivatives for these example functions:

      \dfrac{d}{dx}\left(y \right) &= 1 \cdot \dfrac{dy}{dx} = \dfrac{dy}{dx}\\[8px]\dfrac{d}{dx}\left(y^5 \right) &= 5y^4 \cdot \dfrac{dy}{dx} \\[8px]\dfrac{d}{dx}\left( \sin(y) \right) &= \cos(y) \cdot \dfrac{dy}{dx} \\[8px]\dfrac{d}{dx}\left( \cos(y^2) \right) &= -\sin(y^2) \cdot (2y) \cdot \dfrac{dy}{dx} \\[8px]\dfrac{d}{dx}\left(\ln(y) \right) &= \frac{1}{y} \cdot \dfrac{dy}{dx} \\[8px]\dfrac{d}{dx}\left(\ln(y^5) \right) &= \frac{1}{y^5} \cdot \left(5y^4 \right) \cdot \dfrac{dy}{dx} = \frac{5}{y} \cdot\dfrac{dy}{dx}

      The biggest challenge when learning to do Implicit Differentiation problems is to remember to include this  $\dfrac{dy}{dx}$  term when you take the derivative of something that has a y in it. As always, practicing is the way to learn, and you’ll get good practice problems below.

    • Remember that   $\dfrac{d}{dx}\text{(constant)}= 0$ . (It’s a common error to forget that when doing these problems, even among experts.)
  2. Solve for  $\dfrac{dy}{dx}$.
    • Collect all terms with  $\dfrac{dy}{dx}$  in them on the left side of the equation, all other terms on the right.
    • Factor and divide as necessary to solve for  $\dfrac{dy}{dx}$.

Note: You may use $y’$ instead of  $\dfrac{dy}{dx}$.  They are interchangeable:

$$y’ = \dfrac{dy}{dx}$$

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