Derivatives of $\cos^{-1} x$ and $\tan^{-1} x$
You already know $\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}$. Now complete the trio: derive the derivatives of $\cos^{-1}x$ and $\tan^{-1}x$ from first principles using implicit differentiation, and notice the elegant connection between them. These results unlock a whole family of integration antiderivatives coming in Lessons 13–15.
From Lesson 10, $\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}$. Without looking it up — predict the sign and form of $\dfrac{d}{dx}\cos^{-1}x$. Is it positive or negative? Does $1-x^2$ still appear? Write your reasoning below.
Adding to the derivative of $\sin^{-1}x$, the complete inverse-trig derivative table for HSC Extension 1 is:
These three results are read directly off the HSC reference sheet. You must also be able to derive the $\cos^{-1}x$ and $\tan^{-1}x$ results via implicit differentiation.
- $\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}$ ($|x|<1$)
- $\dfrac{d}{dx}\cos^{-1}x = \dfrac{-1}{\sqrt{1-x^2}}$ ($|x|<1$)
- $\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+x^2}$ (all real $x$)
Key facts
- $\dfrac{d}{dx}\cos^{-1}x = \dfrac{-1}{\sqrt{1-x^2}}$, $|x|<1$
- $\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+x^2}$, all real $x$
- $\dfrac{d}{dx}\sin^{-1}x + \dfrac{d}{dx}\cos^{-1}x = 0$ (since $\sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2}$)
Concepts
- Why the implicit-differentiation method works for inverse functions
- The geometric meaning of the negative sign in $\dfrac{d}{dx}\cos^{-1}x$
- How $1+\tan^2\theta = \sec^2\theta$ produces $\dfrac{1}{1+x^2}$
Skills
- Derive both results using implicit differentiation
- Differentiate expressions involving $\cos^{-1}x$ and $\tan^{-1}x$
- Identify domain restrictions and note when they apply
Let $y = \cos^{-1}x$, so $\cos y = x$ with $y \in [0,\pi]$.
Differentiate both sides with respect to $x$:
Now express $\sin y$ in terms of $x$. Since $y \in [0,\pi]$ we have $\sin y \geq 0$, so:
Substituting back:
$\frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^2}}$; $\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2}$. Both derived via implicit differentiation of the inverse function.
Pause — copy the derivation of $\frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^2}}$ and the key reason for the minus sign (derivative of $\cos$ is $-\sin$, and $\sin y>0$ on $[0,\pi]$) into your book.
Quick check: What is $\dfrac{d}{dx}\cos^{-1}x$?
We just saw that $\frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^2}}$ — identical to the $\sin^{-1}$ result but with a negative sign, arising because $\sin y$ is decreasing on the $\cos^{-1}$ branch $[0,\pi]$. That raises a question: what is the corresponding derivation for $\tan^{-1}x$, and why is the denominator $1+x^2$ rather than $\sqrt{1-x^2}$? This card answers it → $\tan y=x\Rightarrow\sec^2 y\frac{dy}{dx}=1\Rightarrow\frac{dy}{dx}=\frac{1}{1+\tan^2 y}=\frac{1}{1+x^2}$.
Let $y = \tan^{-1}x$, so $\tan y = x$ with $y \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$.
Differentiate both sides with respect to $x$:
Use the identity $\sec^2 y = 1 + \tan^2 y = 1 + x^2$:
Let $y = \tan^{-1}x$, so $\tan y = x$ with $y \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$.
Pause — copy the derivation of $\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2}$: $\tan y=x\Rightarrow\sec^2 y\frac{dy}{dx}=1\Rightarrow\frac{dy}{dx}=\frac{1}{1+x^2}$ into your book.
Did you get this? True or false: the derivation of $\dfrac{d}{dx}\tan^{-1}x$ uses the identity $\sec^2 y = 1 + \tan^2 y$.
Worked examples · 3 in a row, reveal as you go
Find $\dfrac{d}{dx}\left(\cos^{-1}(3x)\right)$.
Differentiate $y = \tan^{-1}(x^2)$.
Differentiate $y = x\tan^{-1}x$.
Fill the gap: $\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+}$ using the identity $\sec^2 y = 1 + \tan^2 y$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $\dfrac{d}{dx}\cos^{-1}x$ and $\dfrac{d}{dx}\sin^{-1}x$ have the same absolute value but opposite signs.
Activities · practice with the ideas
Differentiate $y = \cos^{-1}(2x)$. State the domain restriction.
Find $\dfrac{d}{dx}\tan^{-1}(5x)$.
Differentiate $f(x) = \tan^{-1}\!\left(\dfrac{x}{2}\right)$.
Show, using derivatives, that $\sin^{-1}x + \cos^{-1}x$ is constant.
Differentiate $y = x^2\cos^{-1}x$ using the product rule.
Odd one out: Three of these derivatives are correct. Which one is NOT?
Earlier you predicted whether $\dfrac{d}{dx}\cos^{-1}x$ carries a negative sign.
The exact result is $\dfrac{d}{dx}\cos^{-1}x = \dfrac{-1}{\sqrt{1-x^2}}$ — yes, negative. The key insight: $\cos^{-1}x$ slopes downward everywhere on its domain, so the derivative is always negative. Meanwhile $\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+x^2}$ is always positive (tan⁻¹ slopes upward). Did your intuition match?
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Write down $\dfrac{d}{dx}\tan^{-1}x$. (1 mark)
Q2. Differentiate $y = \cos^{-1}(2x) + \tan^{-1}(2x)$. (2 marks)
Q3. Derive $\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+x^2}$ from first principles using implicit differentiation. (3 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. $\dfrac{-2}{\sqrt{1-4x^2}}$, $|x|<\tfrac{1}{2}$ · 2. $\dfrac{5}{1+25x^2}$ · 3. $\dfrac{1}{2(1+\frac{x^2}{4})} = \dfrac{2}{4+x^2}$ · 4. sum of derivatives $= 0$, confirming constant · 5. $2x\cos^{-1}x - \dfrac{x^2}{\sqrt{1-x^2}}$
Q1 (1 mark): $\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+x^2}$ [1].
Q2 (2 marks): $\dfrac{dy}{dx} = \dfrac{-2}{\sqrt{1-4x^2}} + \dfrac{2}{1+4x^2}$ [1 mark each term, total 2]. Domain note: first term requires $|x|<\tfrac{1}{2}$.
Q3 (3 marks): Let $y = \tan^{-1}x \Rightarrow \tan y = x$ [1]. Differentiate: $\sec^2 y \cdot \dfrac{dy}{dx} = 1$ [1]. Use $\sec^2 y = 1 + \tan^2 y = 1 + x^2$, so $\dfrac{dy}{dx} = \dfrac{1}{1+x^2}$ [1].
Five timed questions on $\cos^{-1}$ and $\tan^{-1}$ derivatives. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering inverse-trig differentiation questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.