Review of the chain rule

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

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 $$\frac {d}{dx} (f(y)) = f'(y) \cdot \frac {d}{dx} (y) = f'(y) \frac {dy}{dx}.$$

Implicit differentiation

The functions we’ve been dealing with so far have been defined explicitly 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, this is not always necessary or even possible to do. Sometimes we choose to or we have to define a function implicitly . In this case, the dependent variable is not stated explicitly in terms of an 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 equations for $y$ and go merrily on your way. However, this can be difficult or even impossible to do. Since we are often faced with a problem of computing derivatives of such functions, we need a method that will enable us to compute derivatives of implicitly defined functions.

We’ll start with a basic example.

Consider the curve (a circle) defined by: $$x^2 + y^2 = 1$$

(a)

Compute $\frac {dy}{dx}$.

(b)

Find the slope of the line tangent to the circle at $\left (\frac {\sqrt {2}}{2},\frac {\sqrt {2}}{2}\right )$.

(c)

Find the slope of the line tangent to the circle at $\left (\frac {\sqrt {2}}{2},-\frac {\sqrt {2}}{2}\right )$.

The curve defined by the equation $x^2 + y^2 = 1$ is not a graph of a function. If we solve for $y$, we obtain two solutions: $y=\sqrt {1-x^2}$ and $y=-\sqrt {1-x^2}$. Therefore, we can say that any point $(x,y)$ on the curve lies on the graph of some function $f$. Starting with $$x^2 + y^2 = 1$$ we differentiate both sides of the equation with respect to $x$ to obtain $$\frac {d}{dx} \left (x^2+y^2\right ) = \frac {d}{dx} 1.$$ Applying the sum rule we see $$\frac {d}{dx} x^2+\frac {d}{dx} y^2 = 0.$$ Let’s examine each of these terms in turn. To start $$\frac {d}{dx} x^2 = \answer [given]{2x}.$$ On the other hand, $\frac {d}{dx} y^2$ is somewhat different. Here we assume that $y = f(x)$ for some function $f$, defined on some open interval (this is true for all points $-1<x<1$). Hence, by the chain rule

$$\begin{align*} \frac{d}{dx} y^2 &= \frac{d}{dx} (f(x))^2 \\ &= 2\cdot f(x) \cdot f'(x) \\ &= 2y\frac{dy}{x}. \end{align*}$$ Putting this together, we are left with the equation $$2x + 2y\frac {dy}{x} =0$$ At this point, we solve for $\frac {dy}{x}$. Write $$\begin{align*} 2x + 2y\frac{dy}{x} &= 0\\ 2y \frac{dy}{x} &= -2x\\ \frac{dy}{x} &= \answer[given]{\frac{-x}{y}}. \end{align*}$$ Remark: Notice that the derivative $\frac {dy}{x}$ is expressed in terms of both variables $x$ and $y$. This should not come as a surprise. If you think about it, the function and its derivative are not determined solely by the value of $x$. Recall what happens if $x=\frac {\sqrt {2}}{2}$.

There are two different points on the curve with the $x-$coordinate $x=\frac {\sqrt {2}}{2}$.

The advantage of the expression for $\frac {dy}{x}$ that we obtained above is that it can be used for computation of the slope of the tangent line at each of these two points, and at any other point on the curve, where defined.

So, for the second part of the problem, we simply plug $x=\frac {\sqrt {2}}{2}$ and $y=\frac {\sqrt {2}}{2}$ into the expression above, hence the slope of the tangent line at this point is $\answer [given]{-1}$. For the third part of the problem, we simply plug $x=\frac {\sqrt {2}}{2}$ and $y=-\frac {\sqrt {2}}{2}$ into the expression 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 $\frac {dy}{dx}$.

(b)

Find the slope of the line tangent to this curve at $(4,2)$.

Starting with $$x^3+y^3 = 9xy,$$ we differentiate both sides of the equation with respect to $x$ to obtain $$\frac {d}{dx} \left (x^3+y^3\right ) = \frac {d}{dx} 9xy.$$ Applying the sum rule we get that $$\frac {d}{dx} x^3+\frac {d}{dx} y^3 = \frac {d}{dx} 9xy.$$ Let’s examine each of these terms in turn. To start $$\frac {d}{dx} x^3 = \answer [given]{3x^2}.$$ On the other hand $\frac {d}{dx} y^3$ is somewhat different. Here you assume that $y = f(x)$, on some interval $I$, and hence by the chain rule $$\begin{align*} \frac{d}{dx} y^3 &= \frac{d}{dx} (f(x))^3 \\ &= 3(f(x))^2 \cdot f'(x) \\ &= 3y^2\frac{dy}{dx}{x}. \end{align*}$$ Considering the final term $\frac {d}{dx} 9xy$, we assume that $y=f(x)$, on some interval $I$. Hence $$\begin{align*} \frac{d}{dx} 9xy &= 9\frac{d}{dx} \Bigl(x\cdot f(x)\Bigr) \\ &= 9 \left(x\cdot f'(x) + f(x)\right)\\ &= 9x \frac{dy}{dx} + 9y. \end{align*}$$ Putting this all together we are left with the equation $$3x^2 + 3y^2\frac {dy}{dx} =9x \frac {dy}{dx} + 9y.$$ At this point, we solve for $\frac {dy}{dx}$. Write $$\begin{align*} 3x^2 + 3y^2\frac{dy}{dx} &= 9x \dfrac{dy}{dx} + 9y\\ 3y^2\frac{dy}{dx} - 9x \frac{dy}{dx} &= 9y - 3x^2\\ \frac{dy}{dx}\left(3y^2-9x\right)&= 9y - 3x^2\\ \frac{dy}{dx} &=\frac{9y - 3x^2}{3y^2-9x} = \frac{3y - x^2}{y^2-3x}. \end{align*}$$

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 $\frac {dy}{dx}$ could sometimes be difficult. In fact, this never happens. All occurrences $\frac {dy}{dx}$ arise from applying the chain rule, and whenever the chain rule is used it deposits a single $\frac {dy}{dx}$ multiplied by some other expression. Hence our expression is linear in $\frac {dy}{dx}$, it will always be possible to group the terms containing $\frac {dy}{dx}$ together and factor out the $\frac {dy}{dx}$, just as in the previous examples.

One more last example: