Chapter 5 · Continuity and Differentiability

Miscellaneous · Q11: Differentiate $x^{x^2-3}+(x-3)^{x^2}$ ⁠, $x>3$

Miscellaneous ExerciseLogarithmic differentiation · Miscellaneous Exercise

The question

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

x^{x^2-3} + (x-3)^{x^2}, \quad \text{for } x > 3.

Exam answer

Let $y = u + v$ ⁠, where $u = x^{x^2 - 3}$ and $v = (x - 3)^{x^2}$ ⁠, $x \gt 3$ ⁠. Then $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ ⁠.

Taking log on both sides of $u = x^{x^2 - 3}$ ⁠,

\log u = (x^2 - 3)\log x

Differentiating both sides w.r.t. $x$ ⁠,

\begin{aligned} \frac1u\frac{du}{dx} &= 2x\log x + \frac{x^2 - 3}{x} \\ \therefore\ \frac{du}{dx} &= x^{x^2 - 3}\Big[\frac{x^2 - 3}{x} + 2x\log x\Big] \end{aligned}

Taking log on both sides of $v = (x - 3)^{x^2}$ ⁠,

\log v = x^2\log(x - 3)

Differentiating both sides w.r.t. $x$ ⁠,

\begin{aligned} \frac1v\frac{dv}{dx} &= 2x\log(x - 3) + \frac{x^2}{x - 3} \\ \therefore\ \frac{dv}{dx} &= (x - 3)^{x^2}\Big[\frac{x^2}{x - 3} \\ &\qquad + 2x\log(x - 3)\Big] \end{aligned}

$\therefore\ \dfrac{dy}{dx} = x^{x^2 - 3}\Big[\dfrac{x^2 - 3}{x} + 2x\log x\Big]$ $+ (x - 3)^{x^2}\Big[\dfrac{x^2}{x - 3} + 2x\log(x - 3)\Big]$ $(x \gt 3)$

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

The one idea this problem is built on

A sum of two variable powers. Logarithms cannot be taken of the whole sum (the log of a sum does not split), so we split first: $y = u + v$ with $u = x^{x^2-3}$ and $v = (x-3)^{x^2}$ ⁠. Each of $u$ and $v$ has a varying base and a varying exponent, so each is handled by logarithmic differentiation. The condition $x > 3$ makes both bases, $x$ and $x - 3$ ⁠, positive.

Given, to find, and the plan

Given: $y = x^{x^2-3} + (x-3)^{x^2}$ for $x > 3$ ⁠.
To find: $\dfrac{dy}{dx}$ ⁠.
Plan: (1) Split into $u + v$ and check both bases are positive. (2) Find $u'$ by logarithms. (3) Find $v'$ by logarithms. (4) Add.

Basics you need

1Sum rule

$(u + v)' = u' + v'$ (rule (1), Section 5.3). Basics: derivative rules

2Logarithms

Defined for positive numbers; $\log p^n = n\log p$ ⁠; $\frac{d}{dx}\log x = \frac1x$ (Section 5.4, Theorem 5). Basics: exponentials and logarithms

3Product and chain rules

$(fg)' = f'g + fg'$ ⁠; $\frac{d}{dx}\log(x - 3) = \frac{1}{x-3}$ by the chain rule. Basics: composite functions

4Logarithmic differentiation

For a power $[f(x)]^{g(x)}$ with $f(x) > 0$ (Section 5.5). Lesson: logarithmic differentiation

Solution

Split the sum; both bases are positive

Let $u = x^{x^2-3}$ and $v = (x-3)^{x^2}$ ⁠, so $y = u + v$ and
$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}.$
For $x > 3$ ⁠: the base $x$ of $u$ is positive, and the base $x - 3$ of $v$ is positive. Hence $u > 0$ and $v > 0$ ⁠, and their logarithms exist.

Taking $\log y$ directly would give $\log(u + v)$ ⁠, which has no useful simplification. Splitting first lets us use logarithms on each power separately. The condition $x > 3$ is there because of the base $x - 3$ ⁠.

Rule (1), Section 5.3; Section 5.5, " $f(x)$ and $u(x)$ must always be positive". Basics: exponentials and logarithms
Differentiate $u = x^{x^2-3}$

Take logarithms: $\log u = (x^2 - 3)\log x$ ⁠. Differentiate both sides (chain rule on the left, product rule on the right):
$\frac1u\frac{du}{dx} = 2x\log x + (x^2-3)\cdot\frac1x.$
Multiply by $u$ ⁠:
$\frac{du}{dx} = x^{x^2-3}\left[\frac{x^2-3}{x} + 2x\log x\right].$
The two factors of the product are $x^2 - 3$ (derivative $2x$ ⁠) and $\log x$ (derivative $\frac1x$ ⁠). The order of the two terms in the bracket does not matter; we wrote them as the key does.

$\log p^n = n\log p$ ⁠, Section 5.4; product rule (2); Theorem 5. Lesson: logarithmic differentiation
Differentiate $v = (x-3)^{x^2}$

Take logarithms: $\log v = x^2\log(x-3)$ ⁠. The factor $\log(x-3)$ has outer function $\log$ and inner function $x - 3$ (derivative 1), so its derivative is $\frac{1}{x-3}$ ⁠. By the product rule:
$\frac1v\frac{dv}{dx} = 2x\log(x-3) + x^2\cdot\frac{1}{x-3}.$
Multiply by $v$ ⁠:
$\begin{aligned} \frac{dv}{dx} = {}&(x-3)^{x^2} \\ &\times\left[\frac{x^2}{x-3} + 2x\log(x-3)\right]. \end{aligned}$
Same method as Step 2. Dividing by $x - 3$ is safe because $x - 3 > 0$ ⁠.

Chain rule, Theorem 4, Section 5.3.1; product rule (2), Section 5.3. Lesson: chain rule
Add the two derivatives
$\begin{aligned} \frac{dy}{dx} = {}&x^{x^2-3} \\ &\times\left[\frac{x^2-3}{x} + 2x\log x\right] \\ &+ (x-3)^{x^2} \\ &\times\left[\frac{x^2}{x-3} + 2x\log(x-3)\right]. \end{aligned}$
Step 1: the derivative of the sum is the sum of the derivatives from Steps 2 and 3.

Rule (1), Section 5.3.

Result

For $x > 3$ ⁠,

\begin{aligned} \frac{dy}{dx} = {}&x^{x^2-3} \\ &\times\left[\frac{x^2-3}{x} + 2x\log x\right] \\ &+ (x-3)^{x^2} \\ &\times\left[\frac{x^2}{x-3} + 2x\log(x-3)\right]. \end{aligned}

(Textbook answer: $x^{x^2-3}\left[\frac{x^2-3}{x} + 2x\log x\right]$ $+ (x-3)^{x^2}\left[\frac{x^2}{x-3} + 2x\log(x-3)\right]$ ⁠, the same.)

Check it with numbers

At $x = 4$ ⁠: $u = 4^{13} = 67108864$ and $v = 1^{16} = 1$ ⁠, so $y = 67108865$ ⁠.

First bracket: $\frac{13}{4} + 8\log 4$ ⁠, about $3.25 + 11.0903549 = 14.3403549$ ⁠. Second bracket: $\frac{16}{1} + 8\log 1 = 16$ ⁠, and $1^{16} = 1$ ⁠. So the formula gives $67108864 \times 14.3403549 + 16$ ⁠, about $962364942$ ⁠.

Difference quotient with $h = 0.000001$ (a small $h$ because the function is so steep): $y(4.000001) \approx 67109827.372$ and $y(3.999999) \approx 67107902.642$ ⁠, so the quotient is about $962364942$ ⁠. ✓

Common mistakes

Taking $\log$ of the whole sum: $\log\left(x^{x^2-3} + (x-3)^{x^2}\right)$ is not $(x^2-3)\log x + x^2\log(x-3)$ ⁠.
Using the power rule on either term; both exponents vary.
Losing the factor $u$ or $v$ when multiplying back: each bracket goes with its own power.

What to take away

Sum of variable powers: split, take logarithms of each term, add.
Each term follows the pattern $\frac{d}{dx}f^g = f^g\left[g'\log f + \frac{g\,f'}{f}\right]$ ⁠, valid where $f > 0$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Sum rule	Steps 1, 4	Rule (1), Section 5.3
$\log$ defined for positive numbers; $\log p^n = n\log p$	Steps 1–3	Section 5.4 ( $\log p^n = n\log p$ is stated there for every real $n$ ⁠, without proof)
$\frac{d}{dx}\log x = \frac1x$	Steps 2, 3	Theorem 5, Section 5.4
Product rule; chain rule	Steps 2, 3	Rule (2), Section 5.3; Theorem 4, Section 5.3.1 (Lesson: chain rule)
Logarithmic differentiation	Steps 2, 3	Section 5.5 (Lesson: logarithmic differentiation)
Answer	Result	Answers, Miscellaneous Exercise on Chapter 5, Q11
Numerical values	Check	Recomputed in the site's verification script

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Miscellaneous · Q11: Differentiate xx2−3+(x−3)x2x^{x^2-3}+(x-3)^{x^2}xx2−3+(x−3)x2⁠, x>3x>3x>3

Basics you need

Solution

Split the sum; both bases are positive

Differentiate u=xx2−3u = x^{x^2-3}u=xx2−3

Differentiate v=(x−3)x2v = (x-3)^{x^2}v=(x−3)x2

Add the two derivatives

Every fact used, and where it comes from

Miscellaneous · Q11: Differentiate $x^{x^2-3}+(x-3)^{x^2}$ ⁠, $x>3$

Differentiate $u = x^{x^2-3}$

Differentiate $v = (x-3)^{x^2}$