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hscscience Ext 1 · Y12
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Module 8 · L12 of 20 ~35 min ⚡ +95 XP available

Chain Rule with Inverse Trig

You can now differentiate $\sin^{-1}x$, $\cos^{-1}x$, and $\tan^{-1}x$. This lesson extends those skills to composite functions — expressions like $\sin^{-1}(ax+b)$, $\tan^{-1}(x^2-3)$, or $\cos^{-1}\!\left(\tfrac{x}{a}\right)$. Mastering the chain rule here is essential groundwork for Lessons 13–15, where these same structures appear as integration results.

Today's hook — You know $\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}$. Without computing it, what do you think $\dfrac{d}{dx}\sin^{-1}(3x+1)$ looks like — is it just the same result with $x$ replaced by $3x+1$, or is there something extra?
0/5QUESTS
01
Recall — gut answer first
+5 XP warm-up

You know $\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}$. Without looking it up — what do you predict $\dfrac{d}{dx}\sin^{-1}(3x+1)$ will be? Is there an extra factor? Where does it come from? Write your reasoning.

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02
The general chain-rule forms
+5 XP to read

When the argument of an inverse-trig function is a function $u = g(x)$ rather than just $x$, the chain rule introduces an extra factor $\dfrac{du}{dx}$:

The three general results are:

  • $\dfrac{d}{dx}\sin^{-1}(g(x)) = \dfrac{g'(x)}{\sqrt{1-(g(x))^2}}$
  • $\dfrac{d}{dx}\cos^{-1}(g(x)) = \dfrac{-g'(x)}{\sqrt{1-(g(x))^2}}$
  • $\dfrac{d}{dx}\tan^{-1}(g(x)) = \dfrac{g'(x)}{1+(g(x))^2}$

These follow directly from applying the chain rule to the basic results from Lessons 10–11.

d/dx f(g(x)) = f'(g(x)) · g'(x) outer deriv (at inner value) × inner deriv f'(u) = 1/√(1−u²) for sin⁻¹ g'(x) = ? inner derivative = g'(x) / √(1 − (g(x))²) sin⁻¹ general form
$\dfrac{d}{dx}\sin^{-1}(g(x)) = \dfrac{g'(x)}{\sqrt{1-(g(x))^2}}$
The extra factor
$\dfrac{d}{dx}\sin^{-1}(3x) = \dfrac{3}{\sqrt{1-9x^2}}$, not $\dfrac{1}{\sqrt{1-9x^2}}$. The factor of $3$ comes from differentiating the inner function $3x$.
Domain update
When applying $\sin^{-1}$ or $\cos^{-1}$ to $g(x)$, the domain condition becomes $|g(x)| < 1$, i.e. $-(g(x))^2 + 1 > 0$. Solve this for $x$.
Preferred notation
In HSC exams you can write the inner function as a single symbol: let $u = g(x)$, differentiate, then substitute $u = g(x)$ back. Show all working for full marks.
03
What you'll master
Know

Key facts

  • $\dfrac{d}{dx}\sin^{-1}(g(x)) = \dfrac{g'(x)}{\sqrt{1-[g(x)]^2}}$
  • $\dfrac{d}{dx}\cos^{-1}(g(x)) = \dfrac{-g'(x)}{\sqrt{1-[g(x)]^2}}$
  • $\dfrac{d}{dx}\tan^{-1}(g(x)) = \dfrac{g'(x)}{1+[g(x)]^2}$
Understand

Concepts

  • Why the inner derivative $g'(x)$ always appears as a numerator factor
  • How to identify the outer and inner functions correctly
  • Why the domain of the derivative narrows when $g(x)$ is not simply $x$
Can do

Skills

  • Differentiate composite inverse-trig functions using the chain rule
  • Apply the chain rule alongside the product rule or quotient rule
  • Determine domain restrictions for composite expressions
04
Key terms
Composite functionA function of the form $f(g(x))$ — "a function of a function." Example: $\sin^{-1}(3x+1)$ is $\sin^{-1}$ applied to $g(x) = 3x+1$.
Outer function $f$The function applied last — in $\sin^{-1}(3x+1)$, the outer function is $\sin^{-1}$.
Inner function $g$The function inside — in $\sin^{-1}(3x+1)$, the inner function is $g(x) = 3x+1$.
Chain rule$\dfrac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$. The outer function is differentiated at the inner value, then multiplied by the inner derivative.
Inner derivative $g'(x)$The derivative of the inner function; becomes a numerator factor in all composite inverse-trig derivatives.
Domain updateFor $\sin^{-1}(g(x))$ and $\cos^{-1}(g(x))$, the domain becomes $|g(x)| < 1$, which must be solved for $x$.
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Applying the chain rule — systematic method
core concept

Use the following four-step method for any composite inverse-trig derivative:

  1. Identify: write $f$ (outer) and $g(x)$ (inner).
  2. Differentiate outer: write $f'(u)$ using the relevant basic result (leaving $u$ as a placeholder).
  3. Differentiate inner: compute $g'(x)$.
  4. Combine: $\dfrac{dy}{dx} = f'(g(x)) \cdot g'(x)$, substituting $g(x)$ for $u$.
$$\frac{d}{dx}\sin^{-1}(ax+b) = \frac{a}{\sqrt{1-(ax+b)^2}}$$
Why the numerator is $g'(x)$ not $1$. With no chain rule, you'd get $\dfrac{1}{\sqrt{1-u^2}}$. The chain rule multiplies by $\dfrac{du}{dx} = g'(x)$, placing $g'(x)$ in the numerator. The denominator is still $\sqrt{1-(g(x))^2}$ — $g(x)$ has replaced $x$ inside the square root.

Chain rule for inverse-trig: identify outer function, write derivative with $u$ in place of $x$, multiply by $\frac{du}{dx}$. Result: $\frac{d}{dx}\tan^{-1}(x/a)=\frac{a}{a^2+x^2}$.

Pause — copy the four-step chain rule method for inverse-trig composites with a worked example of $\frac{d}{dx}\sin^{-1}(3x)$ into your book.

Quick check: What is $\dfrac{d}{dx}\sin^{-1}(3x+1)$?

06
$\tan^{-1}$ with general argument
core concept

We just saw the four-step chain rule method for composite inverse-trig derivatives: (1) identify the outer inverse-trig function; (2) write its derivative with $u$ inside; (3) multiply by $\frac{du}{dx}$; (4) simplify. That raises a question: for $\frac{d}{dx}\tan^{-1}(g(x))$, what is the standard form most useful for the corresponding integral formula in Lesson 14? This card answers it → $\frac{d}{dx}\tan^{-1}(x/a)=\frac{a}{a^2+x^2}$, so $\int\frac{a}{a^2+x^2}\,dx=\tan^{-1}(x/a)+C$.

The standard form used extensively in Lesson 14 (Integrating $\dfrac{a}{a^2+x^2}$) is:

$$\frac{d}{dx}\tan^{-1}\!\left(\frac{x}{a}\right) = \frac{1/a}{1+(x/a)^2} = \frac{1/a}{\tfrac{a^2+x^2}{a^2}} = \frac{a}{a^2+x^2}$$

Equivalently, the reverse result (used for integration) is:

$$\int \frac{a}{a^2+x^2}\,dx = \tan^{-1}\!\left(\frac{x}{a}\right) + C$$
Connection to Lesson 14. This differentiation result is exactly why $\dfrac{1}{a^2+x^2}$ integrates to $\dfrac{1}{a}\tan^{-1}\!\left(\dfrac{x}{a}\right)$. Differentiating to check is always a valid exam strategy.

The standard form used extensively in Lesson 14 (Integrating $\dfrac{a}{a^2+x^2}$) is:

Pause — copy the general form $\frac{d}{dx}\tan^{-1}(x/a)=\frac{a}{a^2+x^2}$ and the integral corollary $\int\frac{a}{a^2+x^2}\,dx=\tan^{-1}(x/a)+C$ into your book.

Did you get this? True or false: $\dfrac{d}{dx}\tan^{-1}\!\left(\dfrac{x}{3}\right) = \dfrac{3}{9+x^2}$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · LINEAR ARGUMENT

Differentiate $y = \cos^{-1}(4x-1)$.

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Outer: $\cos^{-1}$. Inner: $g(x) = 4x-1$, $g'(x) = 4$.
Identify components before applying the formula.
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PROBLEM 2 · NON-LINEAR ARGUMENT

Find $\dfrac{d}{dx}\tan^{-1}(x^2-3)$.

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Inner: $g(x) = x^2-3$, $g'(x) = 2x$.
Differentiate the inner function first.
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PROBLEM 3 · CHAIN + PRODUCT

Differentiate $y = e^x \sin^{-1}(2x)$.

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Product rule: $\dfrac{dy}{dx} = e^x \cdot \sin^{-1}(2x) + e^x \cdot \dfrac{d}{dx}\sin^{-1}(2x)$
Let $u = e^x$, $v = \sin^{-1}(2x)$; $\dfrac{du}{dx} = e^x$.
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Fill the gap: $\dfrac{d}{dx}\sin^{-1}(5x) = \dfrac{$$}{\sqrt{1-25x^2}}$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01
Forgetting the inner derivative
The single most common error: writing $\dfrac{d}{dx}\sin^{-1}(3x) = \dfrac{1}{\sqrt{1-9x^2}}$ instead of $\dfrac{3}{\sqrt{1-9x^2}}$. The chain rule always adds a factor of $g'(x)$ in the numerator — it cannot be forgotten.
Trap 02
Squaring the whole argument vs. just $x$
For $\sin^{-1}(ax+b)$, the denominator is $\sqrt{1-(ax+b)^2}$, NOT $\sqrt{1-a^2x^2}$. Both $a$ and $b$ are inside the square — expand only if asked, or leave factored.
Trap 03
Missing the negative for cos⁻¹
$\dfrac{d}{dx}\cos^{-1}(g(x)) = \dfrac{-g'(x)}{\sqrt{1-g^2}}$. Under exam pressure students often write the positive version. The negative sign applies to all composite $\cos^{-1}$ derivatives, regardless of what $g(x)$ is.

Did you get this? True or false: $\dfrac{d}{dx}\cos^{-1}(3x-2) = \dfrac{3}{\sqrt{1-(3x-2)^2}}$ is correct.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

1

Differentiate $y = \sin^{-1}(5x)$. State the domain.

2

Find $\dfrac{d}{dx}\tan^{-1}(x^2+1)$.

3

Differentiate $f(x) = \cos^{-1}(1-2x)$ and find its domain.

4

Differentiate $y = x^3\tan^{-1}(2x)$ using the product rule.

5

Verify by differentiation that $\displaystyle\int \dfrac{2}{9+x^2}\,dx = \dfrac{2}{3}\tan^{-1}\!\left(\dfrac{x}{3}\right) + C$.

Odd one out: Three of these derivatives are correct. Which one is NOT?

11
Revisit your thinking

Earlier you predicted whether $\dfrac{d}{dx}\sin^{-1}(3x+1)$ has an extra factor.

The exact answer is $\dfrac{3}{\sqrt{1-(3x+1)^2}}$ — yes, the factor of $3$ (the inner derivative) appears in the numerator. This is the chain rule in action: the formula for $\sin^{-1}$ is evaluated at the inner value $3x+1$, then multiplied by $3$. The denominator structure is unchanged — it's just $x$ replaced everywhere by $3x+1$.

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01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 31 mark

Q1. Write down $\dfrac{d}{dx}\tan^{-1}\!\left(\dfrac{x}{a}\right)$ in simplified form. (1 mark)

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ApplyBand 42 marks

Q2. Differentiate $y = \sin^{-1}(3x-1)$ and state the domain of the derivative. (2 marks)

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AnalyseBand 53 marks

Q3. Differentiate $y = \tan^{-1}(x) \cdot \cos^{-1}(x)$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers: 1. $\dfrac{5}{\sqrt{1-25x^2}}$, $|x|<\tfrac{1}{5}$  ·  2. $\dfrac{2x}{1+(x^2+1)^2}$  ·  3. $\dfrac{2}{\sqrt{1-(1-2x)^2}}$, $0<x<1$  ·  4. $3x^2\tan^{-1}(2x) + \dfrac{2x^3}{1+4x^2}$  ·  5. $\dfrac{d}{dx}\left[\dfrac{2}{3}\tan^{-1}\!\left(\dfrac{x}{3}\right)\right] = \dfrac{2}{3}\cdot\dfrac{3}{9+x^2} = \dfrac{2}{9+x^2}$ ✓

Q1 (1 mark): $\dfrac{d}{dx}\tan^{-1}\!\left(\dfrac{x}{a}\right) = \dfrac{a}{a^2+x^2}$ [1].

Q2 (2 marks): $g(x) = 3x-1$, $g'(x) = 3$; $\dfrac{dy}{dx} = \dfrac{3}{\sqrt{1-(3x-1)^2}}$ [1]. Domain: $|3x-1|<1 \Rightarrow \dfrac{0<x<\tfrac{2}{3}}{}$ [1].

Q3 (3 marks): Product rule [1]: $\dfrac{dy}{dx} = \tan^{-1}(x) \cdot \dfrac{-1}{\sqrt{1-x^2}} + \cos^{-1}(x) \cdot \dfrac{1}{1+x^2}$ [1 each term] $= \dfrac{-\tan^{-1}(x)}{\sqrt{1-x^2}} + \dfrac{\cos^{-1}(x)}{1+x^2}$ [1 final form].

01
Boss battle · The Chain Rule Guardian
earn bronze · silver · gold

Five timed questions on composite inverse-trig derivatives. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering chain-rule inverse-trig questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

Module overview · Extension 1 · Checkpoint 3

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