Implicit differentiation

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In this section we differentiate equations without expressing them in terms of a single variable.

Review of the chain rule

Implicit differentiation is really just an application of the chain rule. So recall:

Chain Rule If $f(x)$ and $g(x)$ are differentiable, then $\ddx f(g(x)) = f'(g(x))g'(x).$

Of particular use in this section is the following. If $y$ is a differentiable function of $x$ and if $f$ is a differentiable function, then $\ddx (f(y)) = f'(y) \cdot \ddx (y) = f'(y) \dd [y]{x}.$

Implicit differentiation

The functions we’ve been dealing with so far have been explicit functions, meaning that the dependent variable is written in terms of the independent variable. For example: $y=3x^2-2x+1,\qquad y=e^{3x}, \qquad y = \frac {x-2}{x^2-3x+2}.$ However, there is another type of function, called an implicit function. In this case, the dependent variable is not stated explicitly in terms of the independent variable. Some examples are: $x^2+y^2 = 4,\qquad x^3+y^3 = 9xy, \qquad x^4+3x^2 = x^{2/3}+y^{2/3} = 1.$ Your inclination might be simply to solve each of these for $y$ and go merrily on your way. However this can be difficult and it may require two branches, for example to explicitly plot $x^2+y^2 = 4$ , one needs both $y= \sqrt {4-x^2}$ and $y=-\sqrt {4-x^2}$ . Moreover, it may not even be possible to solve for $y$ . To deal with such situations, we use implicit differentiation. We’ll start with a basic example.

Consider the curve defined by: $x^2 + y^2 = 1$

(a): Compute $\dd [y]{x}$ .
(b): Find the slope of the tangent line at $\left (\frac {\sqrt {2}}{2},\frac {\sqrt {2}}{2}\right )$ .

Starting with $x^2 + y^2 = 1$ we apply the differential operator $\ddx$ to both sides of the equation to obtain $\ddx \left (x^2+y^2\right ) = \ddx 1.$ Applying the sum rule we see $\ddx x^2+\ddx y^2 = 0.$ Let’s examine each of these terms in turn. To start $\ddx x^2 = \answer [given]{2x}.$ On the other hand, $\ddx y^2$ is somewhat different. Here you imagine that $y = f(x)$ , and hence by the chain rule $\begin{align*} \ddx y^2 &= \ddx (f(x))^2 \\ &= 2\cdot f(x) \cdot f'(x) \\ &= 2y\dd[y]{x}. \end{align*}$

Putting this together we are left with the equation $2x + 2y\dd [y]{x} =0$ At this point, we solve for $\dd [y]{x}$ . Write

$\begin{align*} 2x + 2y\dd[y]{x} &= 0\\ 2y \dd[y]{x} &= -2x\\ \dd[y]{x} &= \answer[given]{\frac{-x}{y}}. \end{align*}$

For the second part of the problem, we simply plug $x=\frac {\sqrt {2}}{2}$ and $y=\frac {\sqrt {2}}{2}$ into the formula above, hence the slope of the tangent line at this point is $\answer [given]{-1}$ . We can confirm our results by looking at the graph of the curve and our tangent line: $\graph {x^2+y^2=1,-x+\sqrt {2}}$

Let’s see another illustrative example:

Consider the curve defined by: $x^3+y^3 = 9xy$

(a): Compute $\dd [y]{x}$ .
(b): Find the slope of the tangent line at $(4,2)$ .

Starting with $x^3+y^3 = 9xy,$ we apply the differential operator $\ddx$ to both sides of the equation to obtain $\ddx \left (x^3+y^3\right ) = \ddx 9xy.$ Applying the sum rule we see $\ddx x^3+\ddx y^3 = \ddx 9xy.$ Let’s examine each of these terms in turn. To start $\ddx x^3 = \answer [given]{3x^2}.$ On the other hand $\ddx y^3$ is somewhat different. Here you imagine that $y = f(x)$ , and hence by the chain rule $\begin{align*} \ddx y^3 &= \ddx (y(x))^3 \\ &= 3(f(x))^2 \cdot f'(x) \\ &= 3y^2\dd[y]{x}. \end{align*}$

Considering the final term $\ddx 9xy$ , we again imagine that $y=f(x)$ . Hence

$\begin{align*} \ddx 9xy &= 9\ddx x\cdot f(x) \\ &= 9 \left(x\cdot y'(x) + f(x)\right)\\ &= 9x \dd[y]{x} + 9y. \end{align*}$

Putting this all together we are left with the equation $3x^2 + 3y^2\dd [y]{x} =9x \dd [y]{x} + 9y.$ At this point, we solve for $\dd [y]{x}$ . Write

$\begin{align*} 3x^2 + 3y^2\dd[y]{x} &= 9x \dd[y]{x} + 9y\\ 3y^2\dd[y]{x} - 9x \dd[y]{x} &= 9y - 3x^2\\ \dd[y]{x}\left(3y^2-9x\right)&= 9y - 3x^2\\ \dd[y]{x} &=\frac{9y - 3x^2}{3y^2-9x} = \frac{3y - x^2}{y^2-3x}. \end{align*}$

For the second part of the problem, we simply plug $x=4$ and $y=2$ into the formula above, hence the slope of the tangent line at $(4,2)$ is $\frac {5}{4}$ . We’ve included a plot for your viewing pleasure:

You might think that the step in which we solve for $\dd [y]{x}$ could sometimes be difficult. In fact, this never happens. All occurrences $\dd [y]{x}$ arise from applying the chain rule, and whenever the chain rule is used it deposits a single $\dd [y]{x}$ multiplied by some other expression. Hence our expression is linear in $\dd [y]{x}$ , it will always be possible to group the terms containing $\dd [y]{x}$ together and factor out the $\dd [y]{x}$ , just as in the previous examples.

One more last example:

Consider the curve defined by $e^{xy} - \frac {y}{x} = 4x^2 y^3.$ Compute $\dd [x]{y}$ .

First, notice that the problem asks for $\dd [x]{y}$ , not $\dd [y]{x}$ . So we are considering $x$ as a function of $y$ . This means the variables have changed places! Not to worry, everything is exactly the same. We apply $\dd {y}$ to both sides of the equation to get $\dd {y} \left ( e^{xy} - \frac {y}{x} \right ) = \dd {y} (4x^2 y^3)$ which gives us $-\answer [given]{e^{xy}} \left (y \dd [x]{y} + x \right ) - \frac {x - y \dd [x]{y}}{x^2} = 8xy^3 \dd [x]{y} + 12 x^2 y^2.$ Distributing and multiplying by $x^2$ yields $\begin{align*} -x^2 y e^{xy} \dd[x]{y} &- x^3 e^{xy} - x + y \dd[x]{y}\\ &= 8x^3y^3 \dd[x]{y} + 12x^4y^2. \end{align*}$

Grouping terms, factoring, and dividing finally gives us

$\begin{align*} -x^2 y e^{xy} \dd[x]{y} &+ y \dd[x]{y} - 8x^3y^3 \dd[x]{y} \\ &= x^3 e^{xy} + x + 12x^4 y^2 \end{align*}$

so, $\left ( y - x^2ye^{xy} - 8x^3 y^3 \right ) \dd [x]{y} = x^3 e^{xy}+ x + 12x^4 y^2$ and now we see $\dd [x]{y} = \frac {x^3 e^{xy} + x + 12x^4 y^2}{y - x^2ye^{xy} - 8x^3 y^3}.$