Differentiating Inverse Trigonometric Functions
Introduced| English |
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| inverse trigonometric functions/ɪnˈvɜːs ˌtrɪɡənəʊˈmetrɪk ˈfʌŋkʃnz/ |
Derivatives that hide a square root
- The inverse trig functions — $\arcsin$, $\arctan$, and friends — undo sine, tangent, etc.
- Their derivatives are surprisingly algebraic: no trig at all, just powers and roots.
- Two are worth memorizing outright; the rest follow the same pattern.
- These inverse trigonometric functions 反三角函数 show up in integrals later, so knowing their derivatives pays off twice.
The two you must know
- $\dfrac{d}{dx}[\arcsin x]=\dfrac{1}{\sqrt{1-x^2}}$.
- $\dfrac{d}{dx}[\arctan x]=\dfrac{1}{1+x^2}$.
- Notice: $\arcsin$ has a root in the denominator; $\arctan$ has a clean $1+x^2$.
- Both are always positive on their domains — these inverses are increasing.
The arctangent's gentle slope
y = a·tanh(bx) (arctan-like shape)
$\arctan x$ rises fastest at $0$ (slope $1$) and flattens out — its derivative $\tfrac{1}{1+x^2}$ shrinks toward $0$.
What is $\dfrac{d}{dx}[\arcsin x]$?
$\arcsin$ has the root form $\tfrac{1}{\sqrt{1-x^2}}$.
What is $\dfrac{d}{dx}[\arctan x]$?
$\arctan$ has the clean form $\tfrac{1}{1+x^2}$ (no root).
The rest of the family
- $\dfrac{d}{dx}[\arccos x]=-\dfrac{1}{\sqrt{1-x^2}}$ — same as $\arcsin$ but negative.
- $\dfrac{d}{dx}[\text{arccot}\,x]=-\dfrac{1}{1+x^2}$ — the negative of $\arctan$'s.
- $\dfrac{d}{dx}[\text{arcsec}\,x]=\dfrac{1}{|x|\sqrt{x^2-1}}$.
- The "co-" inverses carry a minus, echoing the ordinary trig pattern.
$\dfrac{d}{dx}[\arccos x]$ is the ____ of $\dfrac{d}{dx}[\arcsin x]$ (same magnitude, opposite sign).
$\arccos$ carries the minus sign.
Select all inverse-trig derivatives that contain a square root.
$\arcsin$, $\arccos$, and arcsec have roots; $\arctan$ is $\tfrac{1}{1+x^2}$ with no root.
Chain them up
- With the chain rule, replace $x$ by an inner function $u$ and multiply by $u'$.
- $\dfrac{d}{dx}[\arctan(3x)]=\dfrac{1}{1+(3x)^2}\cdot 3=\dfrac{3}{1+9x^2}$.
- $\dfrac{d}{dx}[\arcsin(x^2)]=\dfrac{1}{\sqrt{1-(x^2)^2}}\cdot 2x=\dfrac{2x}{\sqrt{1-x^4}}$.
- Same routine as any chain-rule problem: outer formula, then $\times\,u'$.
For $y=\arctan(3x)$, $y'=\dfrac{3}{1+9x^2}$. Find $y'$ at $x=0$.
At $x=0$: $\tfrac{3}{1+0}=3$.
$\dfrac{d}{dx}[\arctan(3x)]=\dfrac{1}{1+9x^2}$ (no inner factor).
The chain rule adds the inner derivative $3$: $\tfrac{3}{1+9x^2}$.
Don't confuse the two shapes: $\arcsin$ gives $\frac{1}{\sqrt{1-x^2}}$ (a root), while $\arctan$ gives $\frac{1}{1+x^2}$ (no root). And with the chain rule, never drop the inner derivative: $\frac{d}{dx}[\arctan(3x)]=\frac{3}{1+9x^2}$, not $\frac{1}{1+9x^2}$.
Differentiate $y=\arctan(x^2)$.
- Outer: $\arctan(u)$ with $u=x^2$, giving $\dfrac{1}{1+u^2}$.
- Inner derivative: $u'=2x$.
- $y'=\dfrac{1}{1+(x^2)^2}\cdot 2x=\dfrac{2x}{1+x^4}$.
Memorize $\frac{d}{dx}[\arcsin x]=\frac{1}{\sqrt{1-x^2}}$ and $\frac{d}{dx}[\arctan x]=\frac{1}{1+x^2}$. The "co-" inverses ($\arccos$, arccot) carry a minus. For composite inverse trigonometric functions, apply the chain rule — the standard formula times the inner derivative $u'$.