Chapter 5 · Continuity and Differentiability

Example 42: $\frac{dy}{dx}$ when $y = a^{t+1/t}$ and $x = \left(t+\frac1t\right)^a$

ExamplesParametric differentiation · Miscellaneous Examples

The question

For a positive constant $a$ find $\dfrac{dy}{dx}$ ⁠, where

y = a^{t + \frac1t}, \quad\text{and}\quad x = \left(t + \frac1t\right)^{a}.

Exam answer

Here $y = a^{t + \frac1t}$ ⁠, $x = \left(t + \frac1t\right)^a$ ⁠, $t > 0$ ⁠.

Differentiating w.r.t. $t$ (chain rule),

\begin{aligned} \frac{dy}{dt} &= a^{t + \frac1t}\log a\cdot\left(1 - \frac{1}{t^2}\right) \\ \frac{dx}{dt} &= a\left(t + \frac1t\right)^{a-1}\left(1 - \frac{1}{t^2}\right) \end{aligned}

For $t \ne 1$ ⁠, $1 - \frac{1}{t^2} \ne 0$ ⁠, and

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

$\therefore\ \dfrac{dy}{dx} = \dfrac{a^{t + \dfrac1t}\log a}{a\left(t + \dfrac1t\right)^{a-1}}$ $(t > 0,\ t \ne 1)$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Both $x$ and $y$ are given in terms of a third variable $t$ ⁠, a parameter. We do not need $y$ as a formula in $x$ ⁠: the chain rule gives $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ wherever $\dfrac{dx}{dt} \ne 0$ ⁠. The two $t$ -derivatives are a contrast worth noticing: in $y$ the base is constant and the exponent varies (an exponential), while in $x$ the exponent is constant and the base varies (a power).

Given, to find, and the plan

Given: $y = a^{t+1/t}$ ⁠, $x = (t + \frac1t)^a$ ⁠, with $a > 0$ a constant.
To find: $\dfrac{dy}{dx}$ ⁠, and the values of $t$ for which it is valid.
Plan: Find where both are defined. Put $s = t + \frac1t$ ⁠. Compute $\frac{dy}{dt}$ (exponential rule plus chain rule) and $\frac{dx}{dt}$ (power rule plus chain rule). Find where $\frac{dx}{dt} = 0$ ⁠. Divide. Check by eliminating $t$ ⁠.

Basics you need

1Parametric differentiation

If $x = f(t)$ and $y = g(t)$ ⁠, then $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ ⁠, provided $\dfrac{dx}{dt} \ne 0$ (Section 5.6). Lesson: parametric differentiation

2Exponential with a constant base

$\frac{d}{ds}(a^s) = a^s\log a$ for a constant $a > 0$ (Example 28, Section 5.5). Basics: exponentials and logarithms

3Power with a constant exponent

$\frac{d}{ds}(s^a) = a s^{a-1}$ for $s > 0$ and any real constant $a$ ⁠. (For $s > 0$ ⁠, $s^a = e^{a\log s}$ ⁠, whose derivative is $e^{a\log s}\cdot\frac{a}{s} = a s^{a-1}$ ⁠.) Basics: derivative rules

4The chain rule

Both $y$ and $x$ are functions of $s = t + \frac1t$ ⁠, which is a function of $t$ (Theorem 4, Section 5.3.1). Lesson: the chain rule

Solution

Where are $x$ and $y$ defined?

$y = a^{t+1/t}$ is defined for every $t \ne 0$ ⁠, since $a > 0$ ⁠. For $x = s^a$ with $s = t + \frac1t$ and a general real exponent $a$ ⁠, we need the base $s > 0$ ⁠. Now
$t + \frac1t = \frac{t^2+1}{t},$
which has the sign of $t$ ⁠. So we work with $t > 0$ ⁠, where $s \ge 2$ (because $t + \frac1t - 2 = \frac{(t-1)^2}{t} \ge 0$ ⁠).

The textbook says both are defined for all real $t \ne 0$ ⁠. That is true for $y$ ⁠, and for $x$ when $a$ is a whole number (any whole-number power of a negative number exists). But a power such as $(-2)^{1/2}$ is not a real number, so for a general positive $a$ we need $t > 0$ ⁠. When $a$ is a whole number, the same calculation below also works for $t < 0$ ⁠.

A real power $s^a$ with arbitrary real $a$ is defined for $s > 0$ as $e^{a\log s}$ ⁠. Basics: exponentials and logarithms
Name the common inner function
$\begin{aligned} s &= t + \frac1t = t + t^{-1}, \\ \frac{ds}{dt} &= 1 - t^{-2} = 1 - \frac{1}{t^2}. \end{aligned}$
Then $y = a^s$ and $x = s^a$ ⁠.

Both $x$ and $y$ are built from the same expression $t + \frac1t$ ⁠. Naming it once means we differentiate it once and use it in both chain rules.

Power rule: $(t^{-1})' = -t^{-2}$ ⁠. Basics: derivative rules
Find $\frac{dy}{dt}$

Outer function $a^s$ (constant base), inner function $s$ ⁠:
$\begin{aligned} \frac{dy}{dt} &= a^s\log a\cdot\frac{ds}{dt} \\ &= a^{t+\frac1t}\left(1 - \frac{1}{t^2}\right)\log a. \end{aligned}$
The base $a$ is fixed and the exponent varies, so this is an exponential function of $s$ ⁠: its derivative is itself times $\log a$ ⁠. The chain rule then multiplies by $\frac{ds}{dt}$ ⁠.

$(a^s)' = a^s\log a$ (Example 28); chain rule (Theorem 4). Example 28
Find $\frac{dx}{dt}$

Outer function $s^a$ (constant exponent), inner function $s$ ⁠, with $s > 0$ ⁠:
$\begin{aligned} \frac{dx}{dt} &= a s^{a-1}\cdot\frac{ds}{dt} \\ &= a\left(t + \frac1t\right)^{a-1}\left(1 - \frac{1}{t^2}\right). \end{aligned}$
Here the exponent $a$ is fixed and the base varies, so the power rule applies: bring the exponent down and lower it by one. Mixing up the two rules (Step 3 and this step) is the main trap in this example.

Power rule for a real exponent and positive base (Basic 3). Basics: derivative rules
Where is $\frac{dx}{dt} \ne 0$ ⁠?

In $\frac{dx}{dt}$ ⁠, the factors $a$ and $(t + \frac1t)^{a-1}$ are positive. So $\frac{dx}{dt} = 0$ exactly when $1 - \frac{1}{t^2} = 0$ ⁠, that is $t^2 = 1$ ⁠, that is $t = \pm 1$ ⁠. For $t > 0$ the only excluded value is $t = 1$ ⁠.

The parametric formula divides by $\frac{dx}{dt}$ ⁠, so we must remove every $t$ that makes it zero. The textbook's " $\frac{dx}{dt} \ne 0$ only if $t \ne \pm1$ " means: $\frac{dx}{dt}$ is non-zero exactly when $t \ne \pm 1$ ⁠.

A product is zero only if one of its factors is zero; a positive number to any real power is positive.
Divide, and cancel the common factor

For $t > 0$ ⁠, $t \ne 1$ ⁠:
$\begin{aligned} \frac{dy}{dx} &= \frac{dy/dt}{dx/dt} \\ &= \frac{a^{t+\frac1t}\left(1-\frac{1}{t^2}\right)\log a}{a\left(t+\frac1t\right)^{a-1}\left(1-\frac{1}{t^2}\right)} \\ &= \frac{a^{t+\frac1t}\log a}{a\left(t+\frac1t\right)^{a-1}}. \end{aligned}$
The factor $1 - \frac{1}{t^2}$ appears on top and bottom and is non-zero (Step 5), so it cancels. The result can also be written $\dfrac{a^{t+\frac1t-1}\log a}{\left(t+\frac1t\right)^{a-1}}$ ⁠, by using $\frac{a^{t+1/t}}{a} = a^{t+1/t-1}$ ⁠.

Section 5.6: $\frac{dy}{dx} = \frac{g'(t)}{f'(t)}$ provided $f'(t) \ne 0$ ⁠. Lesson: parametric differentiation
Check by eliminating $t$

For $t > 0$ ⁠, $s > 0$ and $x = s^a$ ⁠, so $s = x^{1/a}$ and $y = a^{x^{1/a}}$ ⁠, a function of $x$ for $x \ge 2^a$ ⁠. Differentiating directly with the chain rule:
$\begin{aligned} \frac{dy}{dx} &= a^{x^{1/a}}\log a\cdot\tfrac1a\,x^{\frac1a - 1} \\ &= a^s\log a\cdot\tfrac1a\,s^{1-a} = \frac{a^s\log a}{a\,s^{a-1}}, \end{aligned}$
using $x^{\frac1a - 1} = (s^a)^{\frac1a - 1} = s^{1-a}$ ⁠. This is the same answer.

An independent route to the same formula confirms the algebra. It also explains what happens at $t = 1$ ⁠: there $s = 2$ and $x = 2^a$ ⁠, the smallest possible $x$ (the left end of the curve), and the values $t$ and $\frac1t$ give the same point. The parametric method fails at $t = 1$ only because $\frac{dx}{dt} = 0$ there, not because the curve misbehaves.

$(s^a)^{1/a} = s$ for $s > 0$ ⁠; chain rule with $\frac{d}{dx}x^{1/a} = \frac1a x^{\frac1a-1}$ ⁠. Lesson: the chain rule

Result

For $t > 0$ ⁠, $t \ne 1$ (and for all $t \ne 0, \pm1$ when $a$ is a whole number),

\frac{dy}{dx} = \frac{a^{t+\frac1t}\,\log a}{a\left(t + \frac1t\right)^{a-1}}.

(This agrees with the textbook's worked answer.)

Check it with numbers

Take $a = 3$ ⁠, $t = 2$ ⁠, so $s = 2.5$ ⁠. Then $3^{2.5} \approx 15.5884573$ and $\log 3 \approx 1.0986123$ ⁠, and

\begin{aligned} \frac{dy}{dt} &\approx 15.5884573 \times 0.75 \times 1.0986123 \\ &\approx 12.8442530, \\ \frac{dx}{dt} &= 3 \times 2.5^2 \times 0.75 = 14.0625, \end{aligned}

(here $1 - \frac{1}{t^2} = 0.75$ ⁠), so $\frac{dy}{dx} \approx \frac{12.8442530}{14.0625} \approx 0.9133691$ ⁠. The final formula gives $15.5884573 \times 1.0986123$ divided by $3 \times 2.5^2 = 18.75$ ⁠, again about $0.9133691$ ⁠. Directly, $x = 2.5^3 = 15.625$ ⁠, and a difference quotient of $y = 3^{x^{1/3}}$ at $x = 15.625$ also gives $0.91337$ ⁠.

Special case. If $a = 1$ ⁠, then $y = 1^{s} = 1$ is constant, and the formula gives $\log 1 = 0$ ⁠, as it should.

Common mistakes

Swapping the rules. $a^s$ (constant base) needs $(a^s)' = a^s\log a$ ⁠; $s^a$ (constant exponent) needs $(s^a)' = as^{a-1}$ ⁠.
Forgetting $\frac{ds}{dt}$ ⁠. Both derivatives carry the factor $1 - \frac{1}{t^2}$ from the chain rule. It cancels in the end, but only if it was there in both.
Ignoring $\frac{dx}{dt} = 0$ ⁠. At $t = 1$ the quotient $\frac{dy/dt}{dx/dt}$ is $\frac00$ ⁠, so that value must be excluded.
Calling $\frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dx}{dt}$ ⁠. It is a quotient, not a product.

What to take away

Parametric form: differentiate both coordinates with respect to the parameter, then divide, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ ⁠, wherever $\frac{dx}{dt} \ne 0$ ⁠.
Variable in the exponent: use $a^s\log a$ ⁠. Variable in the base: use $as^{a-1}$ ⁠. Variable in both: use logarithms (as in Example 41).
Eliminating the parameter, when it is easy, is the best check.

Every fact used, and where it comes from

Fact	Used in	Source
$s^a = e^{a\log s}$ needs $s > 0$	Step 1	Logarithms only of positive numbers (Section 5.4); the same rewriting as $a^x = e^{x\log a}$ in Example 28, Section 5.5
$(t^n)' = nt^{n-1}$ ⁠, also for real $n$ when $t > 0$	Steps 2, 4, 7	Table 5.3, Section 5.3
$(a^s)' = a^s\log a$	Steps 3, 7	Example 28, Section 5.5
Chain rule	Steps 3, 4, 7	Theorem 4, Section 5.3.1 (Lesson: the chain rule)
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ where $\frac{dx}{dt} \ne 0$	Steps 5, 6	Section 5.6 (Lesson: parametric differentiation)

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Example 42: dydx\frac{dy}{dx}dxdy​ when y=at+1/ty = a^{t+1/t}y=at+1/t and x=(t+1t)ax = \left(t+\frac1t\right)^ax=(t+t1​)a

Basics you need

Solution

Where are xxx and yyy defined?

Name the common inner function

Find dydt\frac{dy}{dt}dtdy​

Find dxdt\frac{dx}{dt}dtdx​

Where is dxdt≠0\frac{dx}{dt} \ne 0dtdx​=0⁠?

Divide, and cancel the common factor

Check by eliminating ttt

Every fact used, and where it comes from

Example 42: $\frac{dy}{dx}$ when $y = a^{t+1/t}$ and $x = \left(t+\frac1t\right)^a$

Where are $x$ and $y$ defined?

Find $\frac{dy}{dt}$

Find $\frac{dx}{dt}$

Where is $\frac{dx}{dt} \ne 0$ ⁠?

Check by eliminating $t$