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4.4: Applications of the Power Chain Rule.

Last updated

Jan 1, 2022
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- 4.3: The generalized power rule.
- 4.5: Some optimization problems.

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The Power Chain Rule

$\text{PCR:} \quad [ u ^{n} (t) ]^{\prime} = n ~ u^{n-1} (t) \times u ^{\prime} (t) = n ~ u^{n-1} (t) ~ u ^{\prime} (t) \label{4.11}$

greatly expands the diversity and interest of problems that we can analyze. Note that the second form omits the $\times$ symbol; multiplication is implied by the juxtaposition of symbols. The general chain rule also will be written

$[ G( u(x) ) ]^{\prime} = G ^{\prime} ( u(x) ) ~ u ^{\prime} (x)$

Some introductory examples follow.

Example 4.4.1

Find the slope of the tangent to the circle, $x ^{2} + y ^{2} = 13$ , at the point (2,3). See Figure $\PageIndex{1}$ .
Also find the slope of the tangent to the circle, $x ^{2} + y ^{2} = 13$ , at the point (3,-2).

$4-3.JPG$

Figure $\PageIndex{1}$ : Graph of the circle $x ^{2} + y ^{2} = 13$ and tangents drawn at the points (2,3) and (3,-2) of the circle.

Solution First check that $2^{2} + 3^{2} = 4 + 9 = 13$ , so that (2,3) is indeed a point of the circle. Then solve for $y$ in $x ^{2} + y ^{2} = 13$ to get

$y_{1} = \sqrt{13 - x ^{2}}$

Then the Power Chain Rule with $n = \frac{1}{2}$ yields

$\begin{array}\ y_{1}^{\prime} &=\left[\sqrt{13-x^{2}}\right]^{\prime}=\left[\left(13-x^{2}\right)^{\frac{1}{2}}\right]^{\prime} & \text { (i) Symbolic identity } \\ &=\frac{1}{2}\left(13-x^{2}\right)^{\frac{1}{2}-1} \times\left[13-x^{2}\right]^{\prime} & \text { (ii) PCR, } n=1 / 2 ~ u=1-x^{2} \\ &=\frac{1}{2}\left(13-x^{2}\right)^{-\frac{1}{2}} \times\left([13]^{\prime}-\left[x^{2}\right]^{\prime}\right) & \text{ (iii) Sum Rule }\\ &=\frac{1}{2}\left(13-x^{2}\right)^{-\frac{1}{2}} \times(0-2 x) & \text{ (iv) Constant and Power Rules }\\ &=-x\left(13-x^{2}\right)^{-\frac{1}{2}} &\\ \end{array}$

Observe the exaggerated ( )'s in steps (iii) and (iv). Students tend to omit writing them, but they are necessary. Without the ( )’s, steps (iii) and (iv) would lead to

$\begin{array}\ y_{1}^{\prime} &=\frac{1}{2}\left(13-x^{2}\right)^{-\frac{1}{2}} \times[13]^{\prime}-\left[x^{2}\right]^{\prime} &\text{(iii) Sum Rule} \\ &=\frac{1}{2}\left(13-x^{2}\right)^{-\frac{1}{2}} \times 0-2 x & \text{(iv) Constant and Power Rules}\\ &=-2 x \end{array}$

a notably simpler answer, but unfortunately incorrect. Always:

Second Notice.

Use parentheses, ( )'s, they are cheap.

To finish the computation, we compute $y ^{\prime} _{1}$ at $x = 2$ and get

\(y ^{\prime} _{1} (2) = (−2) (13 − 2 ^{2})^{- \frac{1}{2}} = - \frac{2}{3}\]

and the slope of the tangent to $x ^{2} + y ^{2} = 13$ at (2,3) is $-2/3$ . An equation of the tangent is

$\frac{y − 3}{x − 2} = - \frac{2}{3} \quad \text{or} \quad y = - \frac{2}{3} x + 4 \frac{1}{3}$

B. Now we find the tangent to the circle $x ^{2} + y ^{2} = 13$ at the point (3,-2). Observe that $3 ^{2} + (-2) ^{2} = 4 + 9 = 13$ so that (2,-3) is a point of $x ^{2} + y ^{2} = 13$ , but (3,-2) does not satisfy

$y_{1} = (13 - x ^{2})^{\frac{1}{2}}$

because

$-2 \neq (13 - (-3)2^{2})^{\frac{1}{2}} = 2$

For (3,-2) we must use the lower semicircle and

$y_{2} = - (13 - x^{2})^{\frac{1}{2}}$

Then

$\begin{array}\ y ^{\prime} _{2} &= -[( 13 - x ^{2}) ^{\frac{1}{2}}] ^{\prime}\\ & = - \frac{1}{2} (13 − x ^{2})^{\frac{1}{2} - 1} [1 - x ^{2}] ^{\prime}\\ & = - \frac{1}{2} (13 − x ^{2})^{- \frac{1}{2}} (−2x)\\ \end{array}$

At $x = 3$

$y ^{\prime} _{2} (3) = - \frac{1}{2} (13 − 3^{2}) ^{- \frac{1}{2}} (-2 \times 3) = \frac{3}{2}$

and the slope of the line drawn is $3/2$ .

It is often helpful to put the denominator of a fraction into the numerator with a negative exponent. For example:

Problem. Compute $P ^{\prime} (t)$ for $P(t) = \frac{5}{(1 + t) ^{2}}$ . Solution

$\begin{aligned} P^{\prime}(t) &=\left[\frac{5}{(1+t)^{2}}\right]^{\prime} \\ &=\left[5(1+t)^{-2}\right]^{\prime} \\ &=5\left[(1+t)^{-2}\right]^{\prime} \\ &=5\left((-2)(1+t)^{-3}\right) \times[(1+t)]^{\prime} \\ &=5(-2)(1+t)^{-3} \times 1=-10(1+t)^{-3} \end{aligned}$

aaa We will find additional important uses of the power rules in the next section.

Exercises for Section 4.4, Applications of the Power Chain Rule.

Exercise 4.4.1 Compute $y ^{\prime} (x)$ for

$y = 2x ^{3} - 5$
$y = \frac{2}{x ^{2}}$
$y = \frac{5}{(x + 1)^{2}}$
$y = (1 + x ^{2}) ^{0.5}$
$y = \sqrt{1 - x ^{2}}$
$y = (1 − x ^{2} ) ^{-0.5}$
$y = (2 − x) ^4$
$y = (3 − x ^{2} ) ^4$
$y = \frac{1}{\sqrt{7 − x ^{2}}}$
$y = (1 + (x - 2)^{2} ) ^2$
$y = (1 + 3x) ^{1.5}$
$y = \frac{1}{\sqrt{16 − x ^{2}}}$
$y = \sqrt{9 - (x - 4)^{2}}$
$y = (9 - x ^{2} ) ^{1.5}$
$y = \sqrt[3] {1 − 3x}$

Exercise 4.4.2 Shown in Figure Ex. 4.4.2 is the ellipse,

$\frac{x ^{2}}{18} + \frac{y ^{2}}{8} = 1$

and a tangent to the ellipse at (3,2).

Find the slope of the tangent.
Find an equation of the tangent.
Find the x- and y-intercepts of the tangent.

$4-4-2.JPG$

Figure for Exercise 4.4.2 Graph of the ellipse $x ^{2}/18 + y ^{2}/8 = 1$ and a tangent to the ellipse at the point (3,2).

Exercise 4.4.3 Shown Figure Ex. 4.4.3 is the ellipse,

$\frac{2x ^{2}}{35} + \frac{3y ^{2}}{35} = 1$

and tangents to the ellipse at (2, 3) and at (4, -1).

Find the slopes of the tangents.
Find equations of the tangents.
Find the point of intersection of the tangents.

$4-4-3.JPG$

Figure for Exercise 4.4.3 Graph of the ellipse $2x ^{2}/35 + 3y ^{2}/35 = 1$ and tangents to the ellipse at the points (2,3) and (4,-1).

Exercise 4.4.4 Shown in Figure Ex. 4.4.4 is the circle with center at (2,3) and radius 5. find the slope of the tangent to this circle at the point (5,7). An equation of the circle is

$(x - 2)^{2} + (y - 3)^{2} = 5^2$

$4-4-4.JPG$

Figure for Exercise 4.4.4 Graph of the circle $(x - 2)^{2} + (y - 3)^{2} = 5^2$ and tangent to the circle at the point (5,7).

Exercises for Section 4.4, Applications of the Power Chain Rule.

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