Chapter 5 · Continuity and Differentiability

Miscellaneous · Q10: Differentiate $x^x+x^a+a^x+a^a$ term by term

Miscellaneous ExerciseLogarithmic differentiation · Miscellaneous Exercise

The question

Differentiate w.r.t. $x$ the function

x^x + x^a + a^x + a^a,

for some fixed $a > 0$ and $x > 0$ ⁠.

Exam answer

Let $y = u + x^a + a^x + a^a$ ⁠, where $u = x^x$ ( $x \gt 0$ ⁠, $a \gt 0$ ⁠).

Taking log on both sides of $u = x^x$ ⁠, $\log u = x\log x$ ⁠. Differentiating w.r.t. $x$ ⁠,

\begin{aligned} \frac1u\frac{du}{dx} &= 1\cdot\log x + x\cdot\frac1x \\ \Rightarrow\ \frac{du}{dx} &= x^x(1 + \log x) \end{aligned}

Also $\frac{d}{dx}(x^a) = ax^{a - 1}$ ⁠, $\frac{d}{dx}(a^x) = a^x\log a$ and $\frac{d}{dx}(a^a) = 0$ ⁠, as $a^a$ is a constant.

$\therefore\ \dfrac{dy}{dx} = x^x(1 + \log x)$ $+ ax^{a - 1} + a^x\log a$ $(x \gt 0)$

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

The one idea this problem is built on

Four terms that look alike but are four different kinds of function. Ask of each term: is the base varying, the exponent varying, both, or neither?

$x^x$ ⁠: both vary (logarithmic differentiation). $x^a$ ⁠: only the base varies (power rule). $a^x$ ⁠: only the exponent varies (exponential rule). $a^a$ ⁠: neither varies, so it is a constant. The derivative of a sum is the sum of the derivatives, so we treat them one at a time.

Given, to find, and the plan

Given: $y = x^x + x^a + a^x + a^a$ ⁠, with $a > 0$ a fixed constant and $x > 0$ ⁠.
To find: $\dfrac{dy}{dx}$ ⁠.
Plan: (1) Split by the sum rule. (2) $x^x$ by logarithmic differentiation. (3) $x^a$ by the power rule. (4) $a^x$ as in Example 28. (5) $a^a$ is constant. (6) Add.

Basics you need

1Sum rule

$(u + v)' = u' + v'$ (rule (1), Section 5.3), and the derivative of a constant is 0 (Class XI, Limits and Derivatives). Basics: derivative rules

2Logarithms

For $p > 0$ ⁠: $\log p^n = n\log p$ (any real $n$ ⁠); $\frac{d}{dx}\log x = \frac1x$ (Theorem 5). Basics: exponentials and logarithms

3Exponential with a constant base

$\frac{d}{dx}a^x = a^x\log a$ for a positive constant $a$ (Example 28).

4Logarithmic differentiation

For $y = [u(x)]^{v(x)}$ with $u(x) > 0$ (Section 5.5). Lesson: logarithmic differentiation

Solution

Split the sum
$\begin{aligned} \frac{dy}{dx} = {}&\frac{d}{dx}(x^x) + \frac{d}{dx}(x^a) \\ &+ \frac{d}{dx}(a^x) + \frac{d}{dx}(a^a). \end{aligned}$
Each term can be differentiated separately. We must not take the logarithm of the whole sum: $\log(p + q)$ does not split into $\log p + \log q$ ⁠, so logarithms only help one term at a time.

Rule (1), Section 5.3. Basics: derivative rules
$x^x$ ⁠: logarithmic differentiation

Let $v = x^x$ ⁠. Since $x > 0$ ⁠, $v > 0$ and $\log v = x\log x$ ⁠. Differentiate both sides (chain rule on the left, product rule on the right):
$\frac1v\frac{dv}{dx} = 1\cdot\log x + x\cdot\frac1x = \log x + 1.$
So $\dfrac{dv}{dx} = x^x(1 + \log x)$ ⁠.

Base and exponent both vary, so neither the power rule nor the exponential rule fits. Logarithms turn the power into the product $x\log x$ ⁠.

Section 5.5; product rule (2), Section 5.3; Theorem 5. Lesson: logarithmic differentiation
$x^a$ ⁠: the power rule

Here the exponent $a$ is a constant. Let $w = x^a$ (positive, as $x > 0$ ⁠). Then $\log w = a\log x$ ⁠, and
$\frac1w\frac{dw}{dx} = \frac{a}{x}, \quad\text{so}\quad \frac{dw}{dx} = \frac{a\,x^a}{x} = a\,x^{a-1}.$
This is the familiar $\frac{d}{dx}x^n = nx^{n-1}$ ⁠. Table 5.3 lists it for the power $n$ ⁠; since $a$ may be any positive real number (not necessarily whole), we confirmed it with logarithms, which works for every real exponent when $x > 0$ ⁠.

Table 5.3; $\log p^n = n\log p$ ⁠, Section 5.4; $\frac{x^a}{x} = x^{a-1}$ (laws of indices). Basics: exponentials and logarithms
$a^x$ ⁠: constant base, variable exponent

Let $z = a^x$ ⁠. Then $\log z = x\log a$ ⁠, where $\log a$ is a constant, so $\frac1z\frac{dz}{dx} = \log a$ and
$\frac{dz}{dx} = a^x\log a.$
This is exactly Example 28 in the book. The derivative of $a^x$ is not $x\,a^{x-1}$ ⁠: that would be the power rule, which needs the exponent to be constant.

Example 28, Section 5.5. Lesson: exponential and logarithmic functions
$a^a$ ⁠: a constant
$\frac{d}{dx}(a^a) = 0.$
$a$ is fixed, so $a^a$ is one fixed number (for instance $2^2 = 4$ when $a = 2$ ⁠); it does not change as $x$ changes.

The derivative of a constant is 0 (Class XI, Limits and Derivatives). Basics: derivative rules
Add the four pieces
$\begin{aligned} \frac{dy}{dx} = {}&x^x(1 + \log x) \\ &+ a\,x^{a-1} + a^x\log a. \end{aligned}$
Step 1 said the derivative of the sum is the sum of the derivatives found in Steps 2 to 5.

Rule (1), Section 5.3.

Result

For $x > 0$ (and fixed $a > 0$ ⁠),

\begin{aligned} \frac{dy}{dx} = {}&x^x(1 + \log x) \\ &+ a\,x^{a-1} + a^x\log a. \end{aligned}

(Textbook answer: $x^x(1 + \log x) + ax^{a-1} + a^x\log a$ ⁠, the same.)

Check it with numbers

Take $a = 2$ and $x = 1$ ⁠. Then $y = 1 + 1 + 2 + 4 = 8$ ⁠, and the formula gives $1(1 + 0) + 2\cdot 1 + 2\log 2$ ⁠, which is about $3 + 1.3862944 = 4.3862944$ ⁠.

Difference quotient with $h = 0.0001$ ⁠: $y(1.0001) \approx 8.00043865$ and $y(0.9999) \approx 7.99956140$ ⁠, so the quotient is about $4.3862944$ ⁠. ✓

Common mistakes

Taking the logarithm of the whole sum and writing $\log y = x\log x + a\log x + \dots$ ⁠. The log of a sum is not the sum of the logs.
Mixing up $x^a$ and $a^x$ : $\frac{d}{dx}x^a = ax^{a-1}$ but $\frac{d}{dx}a^x = a^x\log a$ ⁠.
Giving $a^a$ a non-zero derivative, such as $a\cdot a^{a-1}$ ⁠. It is a constant.
Writing $\frac{d}{dx}x^x = x\cdot x^{x-1}$ : the power rule does not apply when the exponent varies.

What to take away

For any power, first ask which of base and exponent vary: base only (power rule), exponent only ( $a^x\log a$ ⁠), both (logarithms), neither (constant).
Logarithmic differentiation works on a product or a single power, not on a sum: split sums first.

Every fact used, and where it comes from

Fact	Used in	Source
Sum rule; derivative of a constant is 0	Steps 1, 5, 6	Rule (1), Section 5.3 (sum); Class XI, Limits and Derivatives (constant)
$\log p^n = n\log p$ ⁠; $\frac{d}{dx}\log x = \frac1x$	Steps 2–4	Section 5.4 (the log rule is stated there for every real $n$ ⁠, without proof); Theorem 5
Logarithmic differentiation of $x^x$	Step 2	Section 5.5 (Lesson: logarithmic differentiation)
$\frac{d}{dx}x^n = nx^{n-1}$	Step 3	Table 5.3
$\frac{d}{dx}a^x = a^x\log a$	Step 4	Example 28, Section 5.5
Answer	Result	Answers, Miscellaneous Exercise on Chapter 5, Q10
Numerical values	Check	Recomputed in the site's verification script

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Miscellaneous · Q10: Differentiate xx+xa+ax+aax^x+x^a+a^x+a^axx+xa+ax+aa term by term

Basics you need

Solution

Split the sum

xxx^xxx⁠: logarithmic differentiation

xax^axa⁠: the power rule

axa^xax⁠: constant base, variable exponent

aaa^aaa⁠: a constant

Add the four pieces

Every fact used, and where it comes from

Miscellaneous · Q10: Differentiate $x^x+x^a+a^x+a^a$ term by term

$x^x$ ⁠: logarithmic differentiation

$x^a$ ⁠: the power rule

$a^x$ ⁠: constant base, variable exponent

$a^a$ ⁠: a constant