Active Calculus

Figure 2.7.1. At left, the circle given by

x^{2} + y^{2} = 16 .

$x^2 + y^2 = 16\text{.}$ In the middle, the portion of the circle

x^{2} + y^{2} = 16

$x^2 + y^2 = 16$ that has been highlighted in the box at left. And at right, the lemniscate given by

x^{3} - y^{3} = 6 x y .

$x^3 - y^3 = 6xy\text{.}$

Perhaps the simplest and most natural of all such curves are circles. Because of the circle's symmetry, for each

x

$x$ value strictly between the endpoints of the horizontal diameter, there are two corresponding

y

$y$ -values. For instance, in Figure 2.7.1, we have labeled

A = (- 3, \sqrt{7})

$A = (-3,\sqrt{7})$ and

B = (- 3, - \sqrt{7}),

$B = (-3,-\sqrt{7})\text{,}$ and these points demonstrate that the circle fails the vertical line test. Hence, it is impossible to represent the circle through a single function of the form

y = f (x) .

$y = f(x)\text{.}$ But portions of the circle can be represented explicitly as a function of

x,

$x\text{,}$ such as the highlighted arc that is magnified in the center of Figure 2.7.1. Moreover, it is evident that the circle is locally linear, so we ought to be able to find a tangent line to the curve at every point. Thus, it makes sense to wonder if we can compute

\frac{d y}{d x}

$\frac{dy}{dx}$ at any point on the circle, even though we cannot write

y

$y$ explicitly as a function of

x .

$x\text{.}$

We say that the equation

x^{2} + y^{2} = 16

$x^2 + y^2 = 16$ defines

y

$y$ implicitly as a function of

x .

$x\text{.}$ The graph of the equation can be broken into pieces where each piece can be defined by an explicit function of

x .

$x\text{.}$ For the circle, we could choose to take the top half as one function of

x,

$x\text{,}$ namely

y = \sqrt{16 - x^{2}}

$y = \sqrt{16 - x^2}$ and the bottom half as

y = - \sqrt{16 - x^{2}} .

$y = -\sqrt{16 - x^2}\text{.}$ The equation for the circle defines two implicit functions of

x .

$x\text{.}$

The righthand curve in Figure 2.7.1 is called a lemniscate and is just one of many fascinating possibilities for implicitly given curves.

How can we find an equation for

\frac{d y}{d x}

$\frac{dy}{dx}$ without an explicit formula for

y

$y$ in terms of

x ?

$x\text{?}$ The following preview activity reminds us of some ways we can compute derivatives of functions in settings where the function's formula is not known.

Preview Activity 2.7.1.

Let $f$ be a differentiable function of $x$ (whose formula is not known) and recall that $\frac{d}{dx}[f(x)]$ and $f'(x)$ are interchangeable notations. Determine each of the following derivatives of combinations of explicit functions of $x\text{,}$ the unknown function $f\text{,}$ and an arbitrary constant $c\text{.}$

$\frac{d}{dx} \left[ x^2 + f(x) \right]$
$\frac{d}{dx} \left[ x^2 f(x) \right]$
$\frac{d}{dx} \left[ c + x + f(x)^2 \right]$
$\frac{d}{dx} \left[ f(x^2) \right]$
$\frac{d}{dx} \left[ xf(x) + f(cx) + cf(x) \right]$

Subsection 2.7.1 Implicit Differentiation

We begin our exploration of implicit differentiation with the example of the circle given by

x^{2} + y^{2} = 16 .

$x^2 + y^2 = 16\text{.}$ How can we find a formula for

\frac{d y}{d x} ?

$\frac{dy}{dx}\text{?}$

By viewing

y

$y$ as an implicit function of

x,

$x\text{,}$ we think of

y

$y$ as some function whose formula

f (x)

$f(x)$ is unknown, but which we can differentiate. Just as

y

$y$ represents an unknown formula, so too its derivative with respect to

x,

$x\text{,}$

\frac{d y}{d x},

$\frac{dy}{dx}\text{,}$ will be (at least temporarily) unknown.

So we view

y

$y$ as an unknown differentiable function of

x

$x$ and differentiate both sides of the equation with respect to

x .

$x\text{.}$

\frac{d}{d x} [x^{2} + y^{2}] = \frac{d}{d x} [16] .

$\begin{equation*} \frac{d}{dx} \left[ x^2 + y^2 \right] = \frac{d}{dx} \left[ 16 \right]\text{.} \end{equation*}$

On the right, the derivative of the constant

16

$16$ is

0,

$0\text{,}$ and on the left we can apply the sum rule, so it follows that

\frac{d}{d x} [x^{2}] + \frac{d}{d x} [y^{2}] = 0 .

$\begin{equation*} \frac{d}{dx} \left[ x^2 \right] + \frac{d}{dx} \left[ y^2 \right] = 0\text{.} \end{equation*}$

Note carefully the different roles being played by

x

$x$ and

y .

$y\text{.}$ Because

x

$x$ is the independent variable,

\frac{d}{d x} [x^{2}] = 2 x .

$\frac{d}{dx} \left[x^2\right] = 2x\text{.}$ But

y

$y$ is the dependent variable and

y

$y$ is an implicit function of

x .

$x\text{.}$ Recall Preview Activity 2.7.1, where we computed

\frac{d}{d x} [f (x)^{2}] .

$\frac{d}{dx}[f(x)^2]\text{.}$ Computing

\frac{d}{d x} [y^{2}]

$\frac{d}{dx}[y^2]$ is the same, and requires the chain rule, by which we find that

\frac{d}{d x} [y^{2}] = 2 y^{1} \frac{d y}{d x} .

$\frac{d}{dx}[y^2] = 2y^1 \frac{dy}{dx}\text{.}$ We now have that

2 x + 2 y \frac{d y}{d x} = 0 .

$\begin{equation*} 2x + 2y \frac{dy}{dx} = 0\text{.} \end{equation*}$

We solve this equation for

\frac{d y}{d x}

$\frac{dy}{dx}$ by subtracting

2 x

$2x$ from both sides and dividing by

2 y .

$2y\text{.}$

\frac{d y}{d x} = - \frac{2 x}{2 y} = - \frac{x}{y} .

$\begin{equation*} \frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}\text{.} \end{equation*}$

There are several important things to observe about the result that

\frac{d y}{d x} = - \frac{x}{y} .

$\frac{dy}{dx} = -\frac{x}{y}\text{.}$ First, this expression for the derivative involves both

x

$x$ and

y .

$y\text{.}$ This makes sense because there are two corresponding points on the circle for each value of

x

$x$ between

- 4

$-4$ and

4,

$4\text{,}$ and the slope of the tangent line is different at each of these points.

Second, this formula is entirely consistent with our understanding of circles. The slope of the radius from the origin to the point $(a,b)$ is $m_r = \frac{b}{a}\text{.}$ The tangent line to the circle at $(a,b)$ is perpendicular to the radius, and thus has slope $m_t = -\frac{a}{b}\text{,}$ as shown in Figure 2.7.2. In particular, the slope of the tangent line is zero at $(0,4)$ and $(0,-4)\text{,}$ and is undefined at $(-4,0)$ and $(4,0)\text{.}$ All of these values are consistent with the formula $\frac{dy}{dx} = -\frac{x}{y}\text{.}$

Figure 2.7.2. The circle given by

x^{2} + y^{2} = 16

$x^2 + y^2 = 16$ with point

(a, b)

$(a,b)$ on the circle and the tangent line at that point, with labeled slopes of the radial line,

m_{r},

$m_r\text{,}$ and tangent line,

m_{t} .

$m_t\text{.}$

Example 2.7.3.

For the curve given implicitly by $x^3 + y^2 - 2xy = 2\text{,}$ shown in Figure 2.7.4, find the slope of the tangent line at $(-1,1)\text{.}$

Figure 2.7.4. The curve $x^3 + y^2 - 2xy = 2\text{.}$

We begin by differentiating the curve's equation implicitly. Taking the derivative of each side with respect to $x\text{,}$

\begin{equation*} \frac{d}{dx}\left[ x^3 + y^2 - 2xy \right] = \frac{d}{dx} \left[ 2 \right]\text{,} \end{equation*}

by the sum rule and the fact that the derivative of a constant is zero, we have

\begin{equation*} \frac{d}{dx}[x^3] + \frac{d}{dx}[y^2] - \frac{d}{dx}[2xy] = 0\text{.} \end{equation*}

For the three derivatives we now must execute, the first uses the simple power rule, the second requires the chain rule (since $y$ is an implicit function of $x$), and the third necessitates the product rule (again since $y$ is a function of $x$). Applying these rules, we now find that

\begin{equation*} 3x^2 + 2y\frac{dy}{dx} - [2x \frac{dy}{dx} + 2y] = 0\text{.} \end{equation*}

We want to solve this equation for $\frac{dy}{dx}\text{.}$ To do so, we first collect all of the terms involving $\frac{dy}{dx}$ on one side of the equation.

\begin{equation*} 2y\frac{dy}{dx} - 2x \frac{dy}{dx}= 2y - 3x^2\text{.} \end{equation*}

Then we factor the left side to isolate $\frac{dy}{dx}\text{.}$

\begin{equation*} \frac{dy}{dx}(2y - 2x) = 2y - 3x^2\text{.} \end{equation*}

Finally, we divide both sides by $(2y - 2x)$ and conclude that

\begin{equation*} \frac{dy}{dx} = \frac{2y-3x^2}{2y-2x}\text{.} \end{equation*}

Note that the expression for $\frac{dy}{dx}$ depends on both $x$ and $y\text{.}$ To find the slope of the tangent line at $(-1,1)\text{,}$ we substitute the coordinates into the formula for $\frac{dy}{dx}\text{,}$ using the notation

\begin{equation*} \left. \frac{dy}{dx} \right|_{(-1,1)} = \frac{2(1)-3(-1)^2}{2(1)-2(-1)} = -\frac14\text{.} \end{equation*}

This value matches our visual estimate of the slope of the tangent line shown in Figure 2.7.4.

Example 2.7.3 shows that it is possible when differentiating implicitly to have multiple terms involving

\frac{d y}{d x} .

$\frac{dy}{dx}\text{.}$ We use addition and subtraction to collect all terms involving

\frac{d y}{d x}

$\frac{dy}{dx}$ on one side of the equation, then factor to get a single term of

\frac{d y}{d x} .

$\frac{dy}{dx}\text{.}$ Finally, we divide to solve for

\frac{d y}{d x} .

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

We use the notation

{\frac{d y}{d x} |}_{(a, b)}

$\begin{equation*} \left. \frac{dy}{dx} \right|_{(a,b)} \end{equation*}$

to denote the evaluation of

\frac{d y}{d x}

$\frac{dy}{dx}$ at the point

(a, b) .

$(a,b)\text{.}$ This is analogous to writing

f^{'} (a)

$f'(a)$ when

f^{'}

$f'$ depends on a single variable.

There is a big difference between writing

\frac{d}{d x}

$\frac{d}{dx}$ and

\frac{d y}{d x} .

$\frac{dy}{dx}\text{.}$ For example,

\frac{d}{d x} [x^{2} + y^{2}]

$\begin{equation*} \frac{d}{dx}[x^2 + y^2] \end{equation*}$

gives an instruction to take the derivative with respect to

x

$x$ of the quantity

x^{2} + y^{2},

$x^2 + y^2\text{,}$ presumably where

y

$y$ is a function of

x .

$x\text{.}$ On the other hand,

\frac{d y}{d x} (x^{2} + y^{2})

$\begin{equation*} \frac{dy}{dx}(x^2 + y^2) \end{equation*}$

means the product of the derivative of

y

$y$ with respect to

x

$x$ with the quantity

x^{2} + y^{2} .

$x^2 + y^2\text{.}$ Understanding this notational subtlety is essential.

Activity 2.7.2.

Consider the curve defined by the equation $x = y^5 - 5y^3 + 4y\text{,}$ whose graph is pictured in Figure 2.7.5.

Explain why it is not possible to express $y$ as an explicit function of $x\text{.}$
Use implicit differentiation to find a formula for $dy/dx\text{.}$
Use your result from part (b) to find an equation of the line tangent to the graph of $x = y^5 - 5y^3 + 4y$ at the point $(0, 1)\text{.}$
Use your result from part (b) to determine all of the points at which the graph of $x = y^5 - 5y^3 + 4y$ has a vertical tangent line.

Figure 2.7.5. The curve

x = y^{5} - 5 y^{3} + 4 y .

$x = y^5 - 5y^3 + 4y\text{.}$

Hint

Does the graph pass the vertical line test?
Note, for instance, that $\frac{d}{dx}[y^5] = 5y^4\text{.}$
Remember the meaning of $\left. \frac{dy}{dx} \right|_{(0,1)}\text{.}$
What is the slope of a vertical line?

The graph of the curve fails the vertical line test.
$\frac{dy}{dx} = \frac{1}{5y^4 - 15y^2 + 4}\text{.}$
$y = -\frac{1}{6}x + 1\text{.}$
$(1.418697,0.543912)\text{,}$ $(-1.418697,-0.543912)\text{,}$ $(-3.63143, 1.64443)\text{,}$ and $(3.63143, -1.64443)\text{.}$

Because the graph of the curve fails the vertical line test, $y$ cannot be a function of $x\text{.}$ This also confirms our intuition that there is not an algebraic means by which we can rearrange the equation $x = y^5 - 5y^3 + 4y$ to write $y$ in terms of $x\text{.}$
We differentiate implicitly, taking the derivative of each side with respect to $x\text{,}$

\begin{equation*} \frac{d}{dx}[x ]= \frac{d}{dx}[y^5 - 5y^3 + 4y]\text{,} \end{equation*}

and evaluate the elementary derivative on the left and use the sum rule on the right to find that

\begin{equation*} 1 = \frac{d}{dx}[y^5] - \frac{d}{dx}[5y^3] + \frac{d}{dx}[4y]\text{.} \end{equation*}

By the chain and constant multiple rules, viewing $y$ as a function of $x\text{,}$ we now have

\begin{equation*} 1 = 5y^4\frac{dy}{dx} - 15y^2\frac{dy}{dx} + 4\frac{dy}{dx}\text{.} \end{equation*}

Factoring,

\begin{equation*} 1 = \frac{dy}{dx}(5y^4 - 15y^2 + 4)\text{,} \end{equation*}

and therefore

\begin{equation*} \frac{dy}{dx} = \frac{1}{5y^4 - 15y^2 + 4}\text{.} \end{equation*}
To find an equation of the line tangent to the graph of $x = y^5 - 5y^3 + 4y$ at the point $(0, 1)\text{,}$ we only need the slope of the tangent line. Hence we compute

\begin{equation*} \left. \frac{dy}{dx} \right|_{(0,1)} = \frac{1}{5 \cdot 1^4 - 15 \cdot 1^2 + 4} = -\frac{1}{6}\text{.} \end{equation*}

Therefore, the equation of the tangent line is

\begin{equation*} y - 1 = -\frac{1}{6}(x-0) \end{equation*}

or $y = -\frac{1}{6}x + 1\text{.}$
Since a line is vertical whenever its slope is undefined, we seek all points $(x,y)$ that make $\frac{dy}{dx}$ undefined. This will occur precisely when the denominator, $5y^4 - 15y^2 + 4\text{,}$ is zero. Using a graphing utility or computer algebra system to solve the equation $5y^4 - 15y^2 + 4 = 0\text{,}$ we find that this happens at the four approximate $y$-values $y \approx \pm 0.543912, \pm 1.64443\text{.}$ For each such value, we use the original equation $x = y^5 - 5y^3 + 4y$ to find the $x$-value of the point. Doing so, we have established that there are four points at which the tangent line is vertical, and they are approximately $(1.418697,0.543912)\text{,}$ $(-1.418697,-0.543912)\text{,}$ $(-3.63143, 1.64443)\text{,}$ and $(3.63143, -1.64443)\text{.}$

It is natural to ask where the tangent line to a curve is vertical or horizontal. The slope of a horizontal tangent line must be zero, while the slope of a vertical tangent line is undefined. Often the formula for

\frac{d y}{d x}

$\frac{dy}{dx}$ is expressed as a quotient of functions of

x

$x$ and

y,

$y\text{,}$ say

\frac{d y}{d x} = \frac{p (x, y)}{q (x, y)} .

$\begin{equation*} \frac{dy}{dx} = \frac{p(x,y)}{q(x,y)}\text{.} \end{equation*}$

The tangent line is horizontal precisely when the numerator is zero and the denominator is nonzero, making the slope of the tangent line zero. If we can solve the equation

p (x, y) = 0

$p(x,y) = 0$ for either

x

$x$ and

y

$y$ in terms of the other, we can substitute that expression into the original equation for the curve. This gives an equation in a single variable, and if we can solve that equation we can find the point(s) on the curve where

p (x, y) = 0 .

$p(x,y) = 0\text{.}$ At those points, the tangent line is horizontal.

Similarly, the tangent line is vertical whenever

q (x, y) = 0

$q(x,y) = 0$ and

p (x, y) \neq 0,

$p(x,y) \ne 0\text{,}$ making the slope undefined.

Activity 2.7.3.

Consider the curve defined by the equation $y(y^2-1)(y-2) = x(x-1)(x-2)\text{,}$ whose graph is pictured in Figure 2.7.6. Through implicit differentiation, it can be shown that

\frac{d y}{d x} = \frac{(x - 1) (x - 2) + x (x - 2) + x (x - 1)}{(y^{2} - 1) (y - 2) + 2 y^{2} (y - 2) + y (y^{2} - 1)} .

$\begin{equation*} \frac{dy}{dx} = \frac{(x-1)(x-2) + x(x-2) + x(x-1)}{(y^2-1)(y-2) + 2y^2(y-2) + y(y^2-1)}\text{.} \end{equation*}$

Use this fact to answer each of the following questions.

Determine all points $(x,y)$ at which the tangent line to the curve is horizontal. (Use technology appropriately to find the needed zeros of the relevant polynomial function.)
Determine all points $(x,y)$ at which the tangent line is vertical. (Use technology appropriately to find the needed zeros of the relevant polynomial function.)
Find the equation of the tangent line to the curve at one of the points where $x = 1\text{.}$

Figure 2.7.6.

y (y^{2} - 1) (y - 2) = x (x - 1) (x - 2) .

$y(y^2-1)(y-2) = x(x-1)(x-2)\text{.}$

Hint

Note that the numerator of $\frac{dy}{dx}$ is a quadratic function of $x\text{.}$
The denominator of $\frac{dy}{dx}$ is a cubic function of $y\text{.}$
When $x = 1\text{,}$ then $y$ must satisfy the equation $y(y^2-1)(y-2) = 0\text{.}$

Horizontal at $x \approx 0.42265\text{,}$ thus $(0.42265, -1.05782); (0.42265, 0.229478); (0.42265, 0.770522); (0.42265, 2.05782)\text{.}$ There are four more points where $x \approx 1.57735\text{.}$
When $y = \frac{1}{2}, \frac{1 \pm \sqrt{5}}{2}\text{,}$ so one point is $(2.21028, \frac{1}{2})\text{.}$
$y - 1 = \frac{1}{2}(x-1)\text{.}$

To find where the tangent line to the curve is horizontal, we set $\frac{dy}{dx} = 0\text{,}$ which requires that the numerator be zero, or in other words that

\begin{equation*} (x-1)(x-2) + x(x-2) + x(x-1) = 0\text{.} \end{equation*}

Expanding and combining like terms, we find that $3x^2 - 6x + 2 = 0\text{,}$ which occurs where $x = \frac{3\pm\sqrt{3}}{3} \approx 0.42265, 1.57735\text{.}$ From the graph in Figure 2.7.6, we observe that at each such $x$-value, there are several corresponding $y$-values for which the tangent line will be horizontal. For instance, when $x = 0.42265\text{,}$ then $y$ must satisfy the equation

\begin{equation*} y(y^2-1)(y-2) = 0.42265(0.42265-1)(0.42265-2) = 0.384900\text{.} \end{equation*}

Because this is a quartic equation (degree 4) equation in $y\text{,}$ we use computational technology to help us find the solutions. Doing so, we find four approximate values for $y\text{,}$ $y \approx -1.05782, 0.229478, 0.770522, 2.05782\text{,}$ and thus our estimates for four points at which the tangent line is horizontal are

\begin{equation*} (0.42265, -1.05782); (0.42265, 0.229478); (0.42265, 0.770522); (0.42265, 2.05782)\text{.} \end{equation*}

Similar work can be done to find the four points at which the tangent line is horizontal when $x \approx 1.57735\text{.}$
The tangent line to the curve is vertical wherever $\frac{dy}{dx}$ is undefined, which occurs precisely where

\begin{equation*} (y^2-1)(y-2) + 2y^2(y-2) + y(y^2-1) = 0\text{.} \end{equation*}

Expanding and combining like terms, we see that we need to solve the cubic equation $4y^3 - 6y^2 - 2y + 2 = 0\text{;}$ using a computer algebra system, we find that this occurs when $y = \frac{1}{2}, \frac{1 \pm \sqrt{5}}{2}\text{.}$ It now remains to find the $x$-coordinate that corresponds to each such $y$-value. For instance, when $y = \frac{1}{2}\text{,}$ $x$ must satisfy

\begin{equation*} \frac{1}{2}(\frac{1}{4}-1)(\frac{1}{2}-2) = x(x-1)(x-2)\text{,} \end{equation*}

or in other words, $x^3 - 3x^2 + 2x = \frac{9}{16}\text{.}$ Here, too, we use technology to determine that there is only one such $x\text{,}$ and $x \approx 2.21028\text{.}$ Similar work can be done to find the $x$-values that correspond to $y = \frac{1 \pm \sqrt{5}}{2}\text{.}$
There are four points on the curve where $x = 1\text{,}$ which correspond to the $y$-values that satisfy $y(y^2-1)(y-2) = 0\text{:}$ $(1,0)\text{,}$ $(1,1)\text{,}$ $(1,-1)\text{,}$ $(1,2)\text{.}$ We choose the point $(1,1)$ and evaluate $\frac{dy}{dx}$ at this point. Doing so,

\begin{equation*} \left.\frac{dy}{dx} \right|_{(1,1)} = \frac{(1-1)(1-2) + 1(1-2) + 1(1-1)}{(1^2-1)(1-2) + 2\cdot 1^2(1-2) + 1(1^2-1)} = \frac{-1}{-2} = \frac{1}{2}\text{.} \end{equation*}

Thus, the equation of the tangent line to the curve at $(1,1)$ is $y - 1 = \frac{1}{2}(x-1)\text{.}$

Activity 2.7.4.

For each of the following curves, use implicit differentiation to find $dy/dx$ and determine the equation of the tangent line at the given point.

$x^3 - y^3 = 6xy\text{,}$ $(-3,3)$
$\sin(y) + y = x^3 + x\text{,}$ $(0,0)$
$3x e^{-xy} = y^2\text{,}$ $(0.619061,1)$

Hint

Note that $\frac{d}{dx}[6xy]$ requires the product rule.
With $y$ being a function of $x\text{,}$ $\frac{d}{dx}[\sin(y)]$ requires the chain rule.
To calculate $\frac{d}{dx}[x e^{-xy}]\text{,}$ first use the product rule and temporarily defer computing $\frac{d}{dx}[e^{-xy}]\text{.}$

$\frac{dy}{dx}(-3y^2 - 6x) = 6y-3x^2 $ and the tangent line has equation $y - 3 = 1(x+3)\text{.}$
$\frac{dy}{dx} = \frac{3x^2 + 1}{\cos(y) + 1}\text{.}$ and the tangent line has equation $y = \frac{1}{2}x\text{.}$
$\frac{dy}{dx} = \frac{-xye^{-xy} + e^{-xy}}{x^2e^{-xy}+2}\text{.}$ and the tangent line is $y - 1 = 0.110794(x - 0.571433)\text{.}$

Differentiating with respect to $x\text{,}$

\begin{equation*} \frac{d}{dx}[x^3 - y^3] = \frac{d}{dx}[6xy]\text{,} \end{equation*}

so that by the chain and product rules we have

\begin{equation*} 3x^2 - 3y^2 \frac{dy}{dx} = 6x\frac{dy}{dx}+ 6y\text{.} \end{equation*}

Rearranging to get all terms with $\frac{dy}{dx}$ on the same side, it follows that

\begin{equation*} -3y^2 \frac{dy}{dx} - 6x\frac{dy}{dx} = 6y-3x^2\text{,} \end{equation*}

and thus

\begin{equation*} \frac{dy}{dx}(-3y^2 - 6x) = 6y-3x^2\text{.} \end{equation*}

Finally, we have established that

\begin{equation*} \frac{dy}{dx} = \frac{6y-3x^2}{-3y^2 - 6x}\text{,} \end{equation*}

so evaluating at $(-3,3)\text{,}$ we have $\left. \frac{dy}{dx} \right|_{(-3,3)} = \frac{6(3)-3(-3)^2}{-3(3)^2 - 6(-3)} = 1\text{.}$ Thus, the tangent line has equation $y - 3 = 1(x+3)\text{.}$
After differentiating with respect to $x\text{,}$ we have

\begin{equation*} \cos(y) \frac{dy}{dx} + \frac{dy}{dx} = 3x^2 + 1\text{.} \end{equation*}

Taking the usual steps to solve for $\frac{dy}{dx}\text{,}$ we find that

\begin{equation*} \frac{dy}{dx} = \frac{3x^2 + 1}{\cos(y) + 1}\text{.} \end{equation*}

Evaluating the slope of the tangent line at $(0,0)\text{,}$ we have $\left. \frac{dy}{dx} \right|_{(0,0)} = \frac{1}{2}\text{,}$ and thus the tangent line at $(0,0)$ has equation $y = \frac{1}{2}x\text{.}$
When we differentiate both sides with respect to $x\text{,}$

\begin{equation*} \frac{d}{dx}[3x e^{-xy}] = \frac{d}{dx}[y^2]\text{,} \end{equation*}

we first observe that the product rule is needed on the left and the chain rule on the right. Applying those rules, we have

\begin{equation*} 3x\frac{d}{dx}[e^{-xy}] + 3e^{-xy} = 2y\frac{dy}{dx}\text{.} \end{equation*}

Next, we apply the chain rule to differentiate $e^{-xy}\text{,}$ which yields

\begin{equation*} 3xe^{-xy}\frac{d}{dx}[-xy] + 3e^{-xy} = 2y\frac{dy}{dx}\text{.} \end{equation*}

Finally, to complete the process of differentiation, we use the product rule and get

\begin{equation*} 3xe^{-xy}(-x\frac{dy}{dx} - y) + 3e^{-xy} = 2y\frac{dy}{dx}\text{.} \end{equation*}

To solve for $\frac{dy}{dx}\text{,}$ we first expand to have

\begin{equation*} 3x^2e^{-xy}\frac{dy}{dx} - 3xye^{-xy} + 3e^{-xy} = 2y\frac{dy}{dx}\text{,} \end{equation*}

and then the usual algebraic work may be done to deduce that

\begin{equation*} \frac{dy}{dx} = \frac{3xye^{-xy} - 3e^{-xy}}{3x^2e^{-xy}-2y}\text{.} \end{equation*}

Evaluating at the point $(0.571433,1)\text{,}$ it follows that the slope of the tangent line is

\begin{equation*} \left. \frac{dy}{dx} \right|_{(0.571433,1)} = \frac{3 \cdot 0.571433 e^{-0.571433} - 3 e^{-0.571433}}{3(0.571433)^2e^{-0.571433}-2} \approx 0.501835\text{.} \end{equation*}

Thus, the tangent line is given by $y - 1 = 0.501835(x - 0.571433)\text{.}$

Subsection 2.7.2 Summary

In an equation involving $x$ and $y$ where portions of the graph can be defined by explicit functions of $x\text{,}$ we say that $y$ is an implicit function of $x\text{.}$ A good example of such a curve is the unit circle.
We use implicit differentiation to differentiate an implicitly defined function. We differentiate both sides of the equation with respect to $x\text{,}$ treating $y$ as a function of $x$ by applying the chain rule. If possible, we subsequently solve for $\frac{dy}{dx}$ using algebra.
While $\frac{dy}{dx}$ may now involve both the variables $x$ and $y\text{,}$ $\frac{dy}{dx}$ still gives the slope of the tangent line to the curve. It may be used to decide where the tangent line is horizontal ( $\frac{dy}{dx} = 0$ ) or vertical ( $\frac{dy}{dx}$ is undefined), or to find the equation of the tangent line at a particular point on the curve.

3. Implicit differentiation in an equation with inverse trigonometric functions.

4. Slope of the tangent line to an implicit curve.

5. Equation of the tangent line to an implicit curve.

6.

Consider the curve given by the equation $2y^3+y^2-y^5 = x^4 - 2x^3 + x^2\text{.}$ Find all points at which the tangent line to the curve is horizontal or vertical. Be sure to use a graphing utility to plot this implicit curve and to visually check the results of algebraic reasoning that you use to determine where the tangent lines are horizontal and vertical.

Horizontal tangent lines: $(0,-1)\text{,}$ $(0,-0.618)\text{,}$ $(0,1.618)\text{,}$ $(1,-1)\text{,}$ $(1,-0.618)\text{,}$ $(1,1.618)\text{,}$ $(0.5,-1.0493)\text{,}$ $(0.5,0.2104)\text{,}$ $(0.5, 1.6139)\text{.}$ Vertical tangent lines: $(-0.1756,-0.379)\text{,}$ $(0.2912,-0.379)\text{,}$ $(0.7088,-0.379)\text{,}$ $(1.1756,-0.379)\text{,}$ $(-0.8437, 1.235)\text{,}$ and $(1.8437, 1.235)\text{.}$

First, we compute $\frac{dy}{dx}$ by implicit differentiation. Differentiating both sides of the given equation with respect to $x$ while treating $y$ as a function of $x\text{,}$ we see that

\begin{equation*} 6y^2 \frac{dy}{dx} + 2y \frac{dy}{dx} - 5y^4 \frac{dy}{dx} = 4x^3 - 6x^2 + 2x\text{.} \end{equation*}

Removing a factor of $\frac{dy}{dx}$ from the lefthand side and dividing to solve for $\frac{dy}{dx}\text{,}$ we have

\begin{equation*} \frac{dy}{dx} = \frac{4x^3 - 6x^2 + 2x}{6y^2 + 2y - 5y^4} = \frac{2x(2x^2 - 3x + 1)}{y(-5y^3 + 6y + 2)}\text{.} \end{equation*}

The tangent line is potentially horizontal where the numerator of $\frac{dy}{dx}$ is zero and potentially vertical where $\frac{dy}{dx}$ is zero. Solving $2x(2x^2 - 3x + 1) = 0\text{,}$ we see that

\begin{equation*} 2x(2x-1)(x-1) = 0 \end{equation*}

so the numerator is zero at $x = 0, 0.5, 1\text{.}$ The denominator is zero where $y(-5y^3 + 6y + 2) = 0\text{.}$ Using technology to estimate where $-5y^3 + 6y + 2 = 0\text{,}$ we see the denominator is zero at $y = 0$ and $y \approx -0.856, -0.379, 1.235\text{.}$

Next, we need to find the coordinates of these points where potential horizontal or vertical tangent lines occur. First, we consider the potential horizontal tangent line locations. When $x = 0\text{,}$ $2y^3+y^2-y^5 = x^4 - 2x^3 + x^2 = 0\text{,}$ and thus we know $y^2(2y + 1 - y^3) = 0\text{,}$ from which it follows that $y = 0\text{,}$ $y=-1$ or $y \approx -0.618, 1.618\text{.}$ This leads to four points to consider: $(0,0)\text{,}$ $(0,-1)\text{,}$ $(0,-0.618)\text{,}$ and $(0,1.618)\text{.}$ It is apparent from the graph shown below that the tangent line at $(0,0)$ is not horizontal (here both the numerator and denominator of $\frac{dy}{dx}$ are zero), but at each of the other three points, the tangent line is horizontal.

We now reason similarly for the other $x$-values where the numerator is zero. At $x = 1\text{,}$ $2y^3+y^2-y^5 = 1^4 - 2\cdot 1^3 + 1^2 = 0\text{,}$ and here again $y = 0\text{,}$ $y=-1$ or $y \approx -0.618, 1.618\text{.}$ This leads to four potential points where the tangent line is horizontal: $(1,0)\text{,}$ $(1,-1)\text{,}$ $(1,-0.618)\text{,}$ and $(1,1.618)\text{.}$ At $(1,0)\text{,}$ there's not a horizontal tangent line, but at each of the others there is, as seen in the figure. Similar reasoning when $x = 0.5$ shows that $2y^3+y^2-y^5 = 0.5^4 - 2\cdot 0.5^3 + 0.5^2 = 0.625\text{,}$ from which we find that $y \approx -1.0493, 0.2104, 1.6139\text{,}$ and thus the three points $(0.5,-1.0493)\text{,}$ $(0.5,0.2104)\text{,}$ $(0.5, 1.6139)\text{.}$

Returning to where the denominator is zero ($y = 0$ and $y \approx -0.856, -0.379, 1.235$), we now determine points where the tangent line is vertical. Is is apparent from the figure below that there are no vertical tangent lines when $y = 0$ or $y \approx -0.856\text{,}$ so there are two values to pursue: $y \approx -0.379, 1.235\text{.}$

When $y \approx -0.379\text{,}$ it follows that $2(-0.379)^3+(-0.379)^2-(-0.379)^5 = x^4 - 2x^3 + x^2\text{,}$ so

\begin{equation*} x^4 - 2x^3 + x^2 = 0.0426 \end{equation*}

from which we find that $x \approx -0.1756, 0.2912, 0.7088, 1.1756\text{.}$ The figure confirms that there are vertical tangent lines at $(-0.1756,-0.379)\text{,}$ $(0.2912,-0.379)\text{,}$ $(0.7088,-0.379)\text{,}$ and $(1.1756,-0.379)$

Lastly, when $y \approx 1.235\text{,}$ it follows $2(1.235)^3+(1.235)^2-(1.235)^5 = x^4 - 2x^3 + x^2\text{,}$ so

\begin{equation*} x^4 - 2x^3 + x^2 = 2.4195 \end{equation*}

and thus via technology, $x \approx -0.8437, 1.8437\text{.}$ This leads to two additional points at which the tangent line is vertical: $(-0.8437, 1.235)$ and $(1.8437, 1.235)\text{.}$

All of our above work is summarized in the following figure.

7.

For the curve given by the equation $\sin(x+y) + \cos(x-y) = 1\text{,}$ find the equation of the tangent line to the curve at the point $(\frac{\pi}{2}, \frac{\pi}{2})\text{.}$

$y = \frac{\pi}{2} - \left(x-\frac{\pi}{2}\right)\text{.}$

First we need to find the slope of the line tangent to the curve, or $\frac{dy}{dx}\text{.}$ To do so, we differentiate both sides of the equation with respect to $x\text{,}$ treating $y$ as a function of $x$ to obtain

\begin{align*} \frac{d}{dx} \left( \sin(x+y) + \cos(x-y) \right) &= \frac{d}{dx}(1)\\ \cos(x+y) \left(1+\frac{dy}{dx}\right) - \sin(x-y)\left(1-\frac{dy}{dx}\right) &= 0\text{.} \end{align*}

Distributing on the left side of the equation,

\begin{equation*} \cos(x+y) + \cos(x+y)\frac{dy}{dx} - \sin(x-y) + \sin(x-y)\frac{dy}{dx} = 0\text{.} \end{equation*}

Removing a factor of $\frac{dy}{dx}$ from two terms on the right and moving the other two terms to the righthand side,

\begin{equation*} \frac{dy}{dx} \left(\cos(x+y) + \sin(x-y) \right) = \sin(x-y) - \cos(x+y)\text{.} \end{equation*}

Therefore,

\begin{equation*} \frac{dy}{dx} = \frac{\sin(x-y) - \cos(x+y)}{\cos(x+y) + \sin(x-y)}\text{.} \end{equation*}

At the point $\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$ we have that

\begin{equation*} \frac{dy}{dx}\biggm|_{\left(\frac{\pi}{2}, \frac{\pi}{2}\right)} = \frac{\sin(0)-\cos(\pi)}{\cos(\pi) + \sin(0)} = -1\text{.} \end{equation*}

Therefore, the equation of the line tangent to the curve at the point $\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$ is $y = \frac{\pi}{2} - \left(x-\frac{\pi}{2}\right)\text{.}$ A portion of the graph of the curve defined by $\sin(x+y) + \cos(x-y) = 1$ and the tangent line to the curve at the point $\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$ are shown in the following figure.

8.

Implicit differentiation enables us a different perspective from which to see why the rule $\frac{d}{dx} [a^x] = a^x \ln(a)$ holds, if we assume that $\frac{d}{dx}[\ln(x)] = \frac{1}{x}\text{.}$ This exercise leads you through the key steps to do so.

Let $y = a^x\text{.}$ Rewrite this equation using the natural logarithm function to write $x$ in terms of $y$ (and the constant $a$ ).
Differentiate both sides of the equation you found in (a) with respect to $x\text{,}$ keeping in mind that $y$ is implicitly a function of $x\text{.}$
Solve the equation you found in (b) for $\frac{dy}{dx}\text{,}$ and then use the definition of $y$ to write $\frac{dy}{dx}$ solely in terms of $x\text{.}$ What have you found?

$x = \frac{\ln(y)}{\ln(a)}\text{.}$
$1 = \frac{1}{\ln(a)} \cdot \frac{1}{y}\frac{dy}{dx}\text{.}$
$\frac{d}{dx}[a^x] = a^x \ln(a)\text{.}$