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Notes for Introductory Calculus (Math 120)

Brenton LeMesurier

Section 3.8 Implicit Differentiation

References.

In Section 3.7, Derivatives of Inverse Functions, we computed the derivatives of \(y = \ln x\) and \(y = \arcsin x\) functions by using the fact that they are the inverses of the natural exponential and \(\sin\) functions respectively: then we got equations for their values \(y\) in terms of those more familiar functions, \(e^y=x\) and \(\sin y=x\text{.}\) That allowed us to use the Chain Rule to get a formula for \(dy/dx\) in terms of the derivative of those original functions.
These are examples of the strategy of implicit differentiation, where the function to be differentiated is given implicitly as the solution of an equation, rather than by an explicit formula.
Here we will see other uses for this strategy, like computing the slope at a point on a curve when the curve is given by an equation, not as the graph of a known function.
Looking forward, this strategy will be useful in Chapter 4; in particular, Section 4.1
Do this two ways:
  1. By finding an explicit equation \(y=f(x)\) for the curve.
  2. by differentiating the equation
    \begin{equation*} x^2 + [f(x)]^2 = 25 \end{equation*}
    without using a formula for \(f(x)\text{.}\)
Find the tangent line to the point \(P(1,1)\) on the curve \(2x^7+y^7=3xy\text{.}\)

Procedure for Implicit Differentiation, with example \(x^2+xy^2+e^y = 8\).

To find a formula for the derivative of a function \(y=f(x)\) given implicitly by an equation involving \(x\) and \(y\text{:}\)
  1. Differentiate each side of the equation.
    Note that every time \(y\) appears, you must use the Chain Rule.
    \begin{equation*} 2x + \left( y^2 + x2y \cdot \frac{dy}{dx} \right) + e^y \cdot \frac{dy}{dx} = \frac{d}{dx}8 = 0. \end{equation*}
    (The parentheses are around the derivative of \(xy^2\text{,}\) which also requires the product rule.)
  2. Add and subtract to get terms with factor \(\displaystyle\frac{dy}{dx}\) at left, all others at right.
    \begin{equation*} 2xy \cdot \frac{dy}{dx}+ e^y \cdot \frac{dy}{dx} = -2x-y^2. \end{equation*}
  3. Collect the common factor \(\displaystyle\frac{dy}{dx}\) present in every term at left.
    \begin{equation*} \frac{dy}{dx}(2xy + e^y) = -(2x+y^2). \end{equation*}
  4. Divide out to get a formula for \(\displaystyle\frac{dy}{dx}\text{,}\) which is the desired answer.
    \begin{equation*} \frac{dy}{dx} = -\frac{2x+y^2}{2xy + e^y}. \end{equation*}
Note that to use this formula, you need both coordinates of a point of the curve given by the original equation, \(P(x_0,y_0)\text{,}\) not just an \(x\) value.
Hint.
Note that the “hidden” Chain Rule comes up in two places, \(\displaystyle\frac{d}{dx}(y^2)\) and \(\displaystyle\frac{d}{dx}(x+y)\text{.}\)
Solution.
  1. The chain rule at left and product rule at right give:
    \begin{equation*} \cos(x+y)\frac{d}{dx}(x+y) = \frac{d}{dx}(y^2) \cos x + y^2 \frac{d}{dx}(\cos x), \end{equation*}
    so
    \begin{equation*} \cos(x + y) \left(1 + \frac{dy}{dx} \right) = 2 y \frac{dy}{dx} \cos x - y^2 \sin x. \end{equation*}
  2. Moving terms with factor \(\displaystyle\frac{dy}{dx}\) to the left, others to the right,
    \begin{equation*} \frac{dy}{dx}\cos(x+y) -2 y \frac{dy}{dx} \cos x = - y^2 \sin x - \cos(x+y) \end{equation*}
  3. Gathering the common factor \(\displaystyle\frac{dy}{dx}\) at left (and a factor \(-1\) at right),
    \begin{equation*} \frac{dy}{dx}[\cos(x+y) - 2 y \cos x] = - (y^2 \sin x + \cos(x + y)). \end{equation*}
  4. Dividing out,
    \begin{equation*} \frac{dy}{dx} = -\frac{y^2 \sin x + \cos(x + y)}{\cos(x + y) - 2 y \cos x}. \end{equation*}

Higher Derivatives: No Further Implicit Differentiation Needed!

Once a derivative has been found by implicit differentiation, you can compute the second and higher derivatives with no further implicit differentiation:
Solution.
First, implicit differentiation gives \(\displaystyle \frac{dy}{dx} = -\frac{x^3}{y^3}\text{.}\)
Then the second derivative is \(\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx}\frac{dy}{dx} = -\frac{d}{dx}\frac{x^3}{y^3}\text{,}\) and the quotient rule combined with the hidden chain rule result
\begin{equation*} \frac{d(y^3)}{dx}=\frac{d(y^3)}{dy}\frac{dy}{dx} \end{equation*}
gives
\begin{equation*} \frac{d^2y}{dx^2} = - \frac{3x^2 y^3 - x^3 3y^2 dy/dx}{(y^3)^2}. \end{equation*}
Finally, inserting the above result for \(dy/dx\) gives
\begin{equation*} \frac{d^2y}{dx^2} = -\frac{3x^2 y^3 - x^3 3y^2 (-x^3/y^3)}{y^6}, = -\frac{3x^2 y^4 + 3 x^6 }{y^7}\text{.} \end{equation*}

Exercises Exercises

Study Calculus Volume 1, Section 3.8 2 ; in particular Examples 68, 69, 71, 72, both Checkpoint items and Exercises 301, 303, 305, 307, 311, 316, 325, and 329.
openstax.org/books/calculus-volume-1/pages/3-8-implicit-differentiation
openstax.org/books/calculus-volume-1/pages/3-8-implicit-differentiation
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