Implicit Differentiation

The Rules

Explicit versus Implicit Functions

Suppose we have an equation relating two variables: \(x\) and \(y\). If \(y\) is isolated on one side of the equation, we say that \(y\) is explicitly defined as a function of \(x\), or simply, \(y\) is a function of \(x\). If \(y\) is not isolated on one side of the equation, we say that \(y\) is defined implicitly in terms of \(x\), or \(y\) is an implicit function of \(x\).

Explicit Functions

• \(y = x^2 - 3\)

• \(y = \sqrt{x + 7}\)

Implicit Functions

• \(x^2 + xy - y = 1\)

• \(x^3 + y^3 + yx - 4 = 0\)

Computing Derivatives Implicitly

When computing the derivative of a function involving \(x\) and \(y\) (where we treat \(y\) as a function of \(x\)), it may be helpful to replace \(y\) with \(f(x)\) and recall the general power rule:

\[\frac{d}{dx} [f(x)]^n = n[f(x)]^{n-1} f'(x)\]

Example 1

Compute the derivative of \(x^2 y^3\) with respect to \(x\) by treating \(y\) as a function of \(x\).

Step 1:   Replace   \(y\)   with   \(f(x)\).

Since we are treating \(y\) as a function of \(x\), we can rewrite \(x^2 y^3\) as

\[x^2 [f(x)]^3.\]

Step 2:   Differentiate with respect to   \(x\).

Since \(x^2[f(x)]^3\) is a product of two functions of \(x\), namely \(x^2\) and \([f(x)]^3\), we begin by using the product rule.

\[\begin{align*} \frac{d}{dx} x^2 [f(x)]^3 &= \left[\frac{d}{dx}x^2\right][f(x)]^3 + x^2 \left[\frac{d}{dx}[f(x)]^3\right] && \hbox{Product Rule}\\ \\ %&= 2x[f(x)]^3 + x^2 \left[\frac{d}{dx}[f(x)]^3\right] && \hbox{Power \& General Power rules}\\ &= 2x[f(x)]^3 + x^2 3[f(x)]^2f'(x) && \hbox{Power & General Power Rules}\\ %&= 2x[f(x)]^3 + x^2 3[f(x)]^2f'(x) && \hbox{General Power Rule}\\ \end{align*}\]

Step 3:   And finally, replace   \(f(x)\)   with   \(y\)   and   \(f'(x)\)   with   \(y'\).

\[\frac{d}{dx} x^2y^3 = 2xy^3 + 3x^2y^2y'\]

The Method of Implicit Differentiation

Suppose we are given an equation relating the variables \(x\) and \(y\). We can compute the derivative of \(y\) with respect to \(x\) using the following technique.

1. Differentiate both sides of the equation with respect to \(x\).

2. If given, plug in specific values of \(x\) and \(y\) into the resulting equation.

3. Solve for \(y'\) in terms of \(x\) and \(y\).

   1. Put all terms with a factor of \(y'\) on the left side of the equation and everything else on the side of the equation.

   2. Factor out \(y'\) from each term on the left side of the equation.

   3. Divide both sides by the appropriate factor to solve for \(y'\).

Example 2

Find \(\dfrac{dy}{dx}\) where \(y\) is defined implicitly by \({x^2}y - x^4 + y^3 = 1\).

Step 1:   Replace   \(y\)   with   \(f(x)\).

\[x^2f(x) - x^4 + [f(x)]^3 = 1.\]

Step 2:   Differentiate both sides with respect to   \(x\).

\[[2xf(x) + x^2f'(x)] - 4x^3 + 3[f(x)]^2f'(x) = 0 \]

Notice the use of the product rule to compute the derivative of \(x^2f(x)\) and the general power rule to compute the derivative of \([f(x)]^3\).

Step 3:   Rearrange terms so that only terms involving   \(f'(x)\)   are on the left side.

\[x^2f'(x) + 3[f(x)]^2f'(x) = 4x^3 - 2xf(x).\]

Step 4:   Factor out   \(f'(x)\)   on the left side of the equation.

\[(x^2 + 3[f(x)]^2)f'(x) = 4x^3 - 2xf(x).\]

Step 5:   Solve for   \(f'(x)\).

\[f'(x) = \frac{4x^3 - 2xf(x)}{x^2 + 3[f(x)]^2}.\]

Step 6:   And finally, replace   \(f(x)\)   with   \(y\)   and   \(f'(x)\)   with   \(y'\)   (or   \(\frac{dy}{dx}\)).

\[\frac{dy}{dx} = \frac{4x^3 - 2xy}{x^2 + 3y^2}.\]

Observation

Once we get used to the idea of thinking of \(y\) as a function of \(x\), it’s not necessary to replace \(y\) with \(f(x)\). Instead, we can use the general power rule written in the following form:

\[\begin{align*} \frac{d}{dx} y^n &= n y^{n-1} y'\\ &= n y^{n-1} \frac{dy}{dx}. \end{align*}\]

Example 3

Find the equation of the line tangent to the curve defined implicitly by

\[ x^2 y^3 - y^2 + xy = 5 \]

at the point \((2,1)\).

Step 1:   Differentiate both sides with respect to \(x\).

\[\left[ 2xy^3 + x^2 3y^2\frac{dy}{dx}\right] - 2y\frac{dy}{dx} + \left[y + x\frac{dy}{dx}\right] = 0\]

See Implicit Differentiation, Example 1 above for how to differentiate \(x^2y^3\) implicitly.

Step 2:   Plug in specific values of   \(x\)   and   \(y\).

Since we want to find the slope of the line tangent to the curve at the point \((2,1)\), plug in \(x=2\) and \(y=1\).

\[4 + 12\frac{dy}{dx} - 2\frac{dy}{dx} + 1 + 2\frac{dy}{dx} = 0.\]

Step 3:   Simplify and rearrange terms.

\[12\frac{dy}{dx} = -5.\]

Step 4:   Solve for   \(\dfrac{dy}{dx}\).

\[\frac{dy}{dx} = -\frac{5}{12}.\]

Step 5:   Use the point-slope equation of a line:   \(y - b = m(x-a)\).

The equation of the line tangent of the curve at the point \((2,1)\) is

\[y - 1 = -\frac{5}{12}(x-2).\]

Example 4

Find \(\dfrac{dy}{dx}\) at \((2,0)\) where \(y\) is defined implicitly by

\[xy + x = \sqrt{2x + 7y}.\]

Step 1:   Differentiate both sides with respect to   \(x\).

Left Hand Side: $\(\displaystyle \frac{d}{dx}(xy + x) = y + x\frac{dy}{dx} + 1\).$

Right Hand Side:

\[\begin{align*} \displaystyle\frac{d}{dx}\sqrt{2x + 7y} &= \frac{d}{dx}(2x + 7y)^{1/2}\\ &= \frac{1}{2}(2x + 7y)^{-1/2}\frac{d}{dx}(2x + 7y)\\ &= \frac{1}{2\sqrt{2x + 7y}}\left(2 + 7\frac{dy}{dx}\right). \end{align*}\]

Therefore,

\[y + x\frac{dy}{dx} + 1 = \frac{1}{2\sqrt{2x + 7y}}\left(2 + 7\frac{dy}{dx}\right).\]

Step 2:   Plug in the values   \(x=2\)   and   \(y=0\).

\[2\frac{dy}{dx} + 1 = \frac{1}{4}\left(2 + 7\frac{dy}{dx}\right).\]

Step 3:   Expand both sides of the equation so that each side is written as a sum.

\[2\frac{dy}{dx} + 1 = \frac{1}{2} + \frac{7}{4}\frac{dy}{dx}.\]

Step 4:   Rearrange terms.

Subtract \(1\) from both sides and subtract \(\dfrac{7}{4}\dfrac{dy}{dx}\) from both sides.

\[\frac{1}{4}\frac{dy}{dx} = -\frac{1}{2}.\]

Step 5:   Solve for   \(\dfrac{dy}{dx}\).

Multiply both sides by \(4\):

\[\frac{dy}{dx} = -2.\]

Observation

Alternatively, we could multiply both sides of the equation in Step 2 by \(4\) to get

\[8\frac{dy}{dx} + 4 = 2 + 7\frac{dy}{dx}.\]

Now rearranging terms is a little easier since we don’t have to deal with any fractions.

Example 5

The demand equation for Dr. Hager’s Make Math Great Again video series is given by

\[p = -4x^2 - 2x + 30\]

where \(p\) is the wholesale unit price in dollars and \(x\) is the quantity demanded in units of a thousand. Compute the elasticity of demand when \(x=2\).

Observation

Ordinarily when we compute elasticity of demand, the first thing we do is find \(x = f(p)\) by solving for \(x\) in terms of \(p\) in the given demand equation. However in this case, solving for \(x\) is easier said than done. So instead, we will use implicit differentiation to compute \(f'(p)\) when the time comes.

Step 1:   Compute the unit price   \(p\)   when   \(x=2\).

\[\begin{align*} p &= -4(2^2) - 2(2) + 30 & \hbox{since $p = -4x^2 - 2x + 30$} \\ &= -16 - 4 + 30 \\ &= 10 \end{align*}\]

Step 2:   Compute   \(f(10)\).

Recall that \(x=f(p)\) is the quantity demanded (in thousands) when the unit price is \(p\) dollars. We were given \(x=2\) (and used that to determine \(p=10\) in Step 1), which is equivalent to saying \(f(10) = 2\).

Step 3:   Compute   \(f'(10)\)

Compute the derivative of both sides of the demand equation (\(p = -4x^2 - 2x + 30\)) with respect to \(p\). Here is where we use implicit differentiation by thinking of \(x\) as a function of \(p\) (i.e., \(x=f(p)\)).

\[\begin{align*} 1 &= \frac{d}{dp}(-4x^2 - 2x + 30) && \hbox{since $\dfrac{d}{dp}p = 1$}\\ &= -8xx' - 2x' \end{align*}\]

Now plug in \(x=2\) to get

\[1 = -16x' - 2x' = -18x'\]

and therefore \(x' = f'(10) = -1/18\).

Step 4:   Compute elasticity of demand.

Compute elasticity of demand using the values \(p=10\), \(f(10) = 2\), and \(f'(10) = -1/18\).

\[\begin{align*} E(10) &= -\frac{10\cdot (-1/18)}{2} && \hbox{since $E(p) = -\dfrac{p\cdot f'(p)}{f(p)}$} \\ &= \frac{10}{2}\cdot \frac{1}{18} \\ &= \frac{5}{18} \end{align*}\]