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Calculus BC · Unit 3 · 3.4
AP Calculus · Differentiation: Composite, Implicit, and Inverse Functions
Differentiating Inverse Trigonometric Functions
Six results, three to memorise — and every one of them is algebraic.
key idea$\left(\arcsin x\right)'=\frac{1}{\sqrt{1-x^{2}}};\qquad \left(\arctan x\right)'=\frac{1}{1+x^{2}}$
The rule
Six results, but only three to learn — the co- versions are the same expression with a minus sign, exactly as with the ordinary trig derivatives. Notice that every one is an algebraic function: no trig survives in the answer.
| $f(x)$ | $f'(x)$ | Co- partner | Domain of $f$ |
|---|
| $\arcsin x$ | $\dfrac{1}{\sqrt{1-x^{2}}}$ | $\arccos x\to-\dfrac{1}{\sqrt{1-x^{2}}}$ | $[-1,1]$ |
| $\arctan x$ | $\dfrac{1}{1+x^{2}}$ | $\text{arccot }x\to-\dfrac{1}{1+x^{2}}$ | all reals |
| $\text{arcsec }x$ | $\dfrac{1}{\lvert x\rvert\sqrt{x^{2}-1}}$ | $\text{arccsc }x\to-\dfrac{1}{\lvert x\rvert\sqrt{x^{2}-1}}$ | $\lvert x\rvert\ge1$ |
Three rows, six results. The arctan derivative is the one that appears most often — it is the integrand behind $\arctan$ antiderivatives.
Pattern$1-x^{2}$ for arcsin, $1+x^{2}$ for arctan — the sign distinguishes them
No trig in the answerinverse-trig derivatives are all algebraic
Co- means minussame expression, negated — nothing new to memorise
Pattern recognition
recognising the forms
- $\sqrt{1-x^{2}}$ under a fraction bar ⇒ think arcsin.
- $1+x^{2}$ in a denominator ⇒ think arctan.
- $\arcsin$ is only defined on $[-1,1]$, so its derivative is too.
- Any argument other than a bare $x$ needs the chain rule.
Worked examples
EX 1
Derive one implicitly
$\arcsin x$Show that $\dfrac{d}{dx}[\arcsin x]=\dfrac{1}{\sqrt{1-x^{2}}}$.
1
Rewrite without the inverse
Let $y=\arcsin x$, so $\sin y=x$ with $y\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$.
2
Differentiate implicitly
$\cos y\dfrac{dy}{dx}=1$, so $\dfrac{dy}{dx}=\dfrac{1}{\cos y}$.
3
Express $\cos y$ in terms of $x$
From $\sin^{2}y+\cos^{2}y=1$: $\cos y=\sqrt{1-x^{2}}$.
4
Justify the positive root
On $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ cosine is non-negative, so the positive root is correct.
5
Conclude
$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^{2}}}$. The restricted range is what makes the sign unambiguous.
EX 2
With the chain rule
$\arctan(3x)$Differentiate $y=\arctan(3x)$.
1
Identify the layers
Outer $\arctan u$, inner $u=3x$.
2
Differentiate the outer
$\dfrac{1}{1+u^{2}}$ becomes $\dfrac{1}{1+(3x)^{2}}=\dfrac{1}{1+9x^{2}}$.
3
Multiply by the inner derivative
$\dfrac{d}{dx}[3x]=3$.
4
Assemble
$y'=\dfrac{3}{1+9x^{2}}$.
5
Guard the square
$(3x)^{2}=9x^{2}$, not $3x^{2}$. Squaring the whole inner function is where this goes wrong.
EX 3
Product with an inverse trig factor
$x\arcsin x$Differentiate $y=x\arcsin x$.
1
Recognise a product
$f=x$ and $g=\arcsin x$.
2
Differentiate each
$f'=1$ and $g'=\dfrac{1}{\sqrt{1-x^{2}}}$.
3
Apply the product rule
$y'=\arcsin x+\dfrac{x}{\sqrt{1-x^{2}}}$.
4
State the domain
$\arcsin$ needs $\lvert x\rvert\le1$, and the derivative additionally needs $\lvert x\rvert\lt 1$ so the root is non-zero.
5
Note the mixed form
The answer contains both an inverse-trig term and an algebraic one. That is normal and cannot be simplified further.
EX 4
Recognise it backwards
antiderivativeEvaluate $\displaystyle\int\dfrac{dx}{1+25x^{2}}$.
1
Spot the arctan form
A denominator $1+(\text{something})^{2}$ signals $\arctan$.
2
Match the inner function
$25x^{2}=(5x)^{2}$, so the inner function is $u=5x$.
3
Account for the chain factor
$\dfrac{d}{dx}\left[\arctan(5x)\right]=\dfrac{5}{1+25x^{2}}$ — five times what we want.
4
Divide it out
$\displaystyle\int\dfrac{dx}{1+25x^{2}}=\dfrac{1}{5}\arctan(5x)+C$.
5
Why this matters
These derivative facts are used far more often in reverse, so learning to recognise the forms pays twice.
Watch out
$1-x^{2}$ versus $1+x^{2}$
Arcsine has a minus under a square root; arctangent has a plus and no root. Swapping them is the commonest error here.
$\sin^{-1}x$ is not $\dfrac{1}{\sin x}$
The $-1$ marks the inverse function. The reciprocal of sine is $\csc x$, a completely different function with a different derivative.
The chain factor squares too
For $\arctan(3x)$ the denominator is $1+9x^{2}$, not $1+3x^{2}$. Square the entire inner function.
Mind the domain
$\arcsin$ and $\arccos$ live on $[-1,1]$, and their derivatives on $(-1,1)$ — at the endpoints the tangent is vertical.