AP Calculus AB/BC · Unit 3 · Topic 3.4
Differentiating Inverse Trigonometric Functions
Apply inverse trig derivative formulas and chain-rule factors.
1. Topic Focus
Differentiate nested, implicit, inverse, and higher-order relationships by choosing procedures that match the function structure.
This topic: Apply inverse trig derivative formulas and chain-rule factors.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
For arctan(3x), the derivative is 3/(1+9x²).
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
Trigonometric functions are periodic, so they must be restricted to one-to-one intervals before inverses can be defined. An inverse trigonometric function takes a ratio as input and returns a principal angle. For example,
The notation \(\sin^{-1}x\) means \(\arcsin x\), not \(1/\sin x\). Reciprocal trigonometric functions such as cosecant are different functions.
Deriving the inverse-sine formula. Differentiate \(\sin y=x\) implicitly:
On the principal range of arcsine, \(\cos y\ge0\). Since \(\cos^2y=1-\sin^2y=1-x^2\),
The function \(\arcsin x\) is defined at \(x=\pm1\), but its finite derivative is not; the graph has vertical tangents there.
Deriving the inverse-tangent formula. If \(y=\arctan x\), then \(\tan y=x\). Implicit differentiation and \(\sec^2y=1+\tan^2y\) give
| Function | Derivative | Derivative exists when |
|---|---|---|
| \(\arcsin x\) | \(1/\sqrt{1-x^2}\) | \(|x|<1\) |
| \(\arccos x\) | \(-1/\sqrt{1-x^2}\) | \(|x|<1\) |
| \(\arctan x\) | \(1/(1+x^2)\) | All real \(x\) |
| \(\operatorname{arccot}x\) | \(-1/(1+x^2)\) | All real \(x\) |
| \(\operatorname{arcsec}x\) | \(1/(|x|\sqrt{x^2-1})\) | \(|x|>1\) |
| \(\operatorname{arccsc}x\) | \(-1/(|x|\sqrt{x^2-1})\) | \(|x|>1\) |
The arccotangent convention shown is the standard calculus convention with a decreasing principal branch. For arcsecant and arccosecant, the absolute value in the denominator is essential.
Composite-input formulas. Replace \(x\) by \(u(x)\) everywhere in the outer formula, then multiply by \(u'(x)\):
For example, the denominator of \((\arctan(3x))'\) is \(1+(3x)^2\), while the numerator contains the inner derivative 3.
Domain routine. First find where the original inverse-trigonometric expression is defined. Then check where its derivative formula is finite. For \(\arcsin(u)\) and \(\arccos(u)\), the function permits \(|u|\le1\), but the displayed derivative requires \(|u|<1\). For arcsecant and arccosecant, the functions permit \(|u|\ge1\), while finite derivatives require \(|u|>1\).
Sign and graph checks. Arcsine and arctangent are increasing, so their derivatives are positive wherever they exist. Arccosine is decreasing, so its derivative is negative. These graph facts are useful for detecting a lost minus sign.
6. Detailed Worked Example and Error Check
Example 1: Linear input to arcsine. Let \(y=\arcsin(3x-1)\). Then
The function is defined when \(0\le x\le2/3\), while the derivative is finite only for \(0<x<2/3\).
Example 2: Arccosine of a power.
The original function has real domain \([-1,1]\), and the displayed derivative exists for \(-1<x<1\). The negative sign belongs to the outer arccosine derivative.
Example 3: Arctangent of an exponential.
Because arctangent accepts every real input and the denominator is always positive, this derivative exists for every real \(x\).
Example 4: Product rule with an inverse function. For \(F(x)=x^2\arcsin x\),
The product rule creates two terms; only the second term differentiates arcsine.
Example 5: Tangent line. For \(y=\arcsin x\) at \(x=1/2\), the point is \((1/2,\pi/6)\) and
Thus the tangent line is \(y-\frac{\pi}{6}=\frac2{\sqrt3}(x-\frac12)\).
Example 6: Arcsecant and the absolute value.
for \(|x|>1/2\). Replacing \(|x|\) by \(x\) would produce the wrong sign on the negative branch.
Example 7: Second derivative and concavity. If \(y=\arctan x\), then
Therefore arctangent is concave up for \(x<0\), concave down for \(x>0\), and changes concavity at the origin.
AP error check. Do not confuse inverse and reciprocal notation, omit the inner derivative, forget the negative sign for arccosine, write \(1-u^2\) instead of \(1+u^2\) for arctangent, drop the absolute value from the arcsecant formula, or report endpoint values as finite derivatives.
7. AP Reasoning Routine
Mark inner and outer functions, track every derivative factor, solve algebraically for the requested derivative, and verify the result's domain.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Differentiate and state the real derivative domain when requested.
(a) Differentiate \(y=\arcsin(2x)\).
(b) Differentiate \(y=\arccos(1-3x)\).
(c) Differentiate \(y=\arctan(x^2)\).
(d) Differentiate \(y=x\arcsin x\).
(e) Find the tangent line to \(y=\arctan x\) at \(x=1\).
(f) Differentiate \(y=\arcsin(x^2-2)\), then state the domain of the function and the domain of its derivative.
(g) Starting from \(y=\arcsin x\), use implicit differentiation to derive its derivative formula.
(h) Differentiate \(y=\operatorname{arcsec}(3x)\) and state where the derivative exists.
Check the solution
(a) \(y'=2/\sqrt{1-4x^2}\), defined for \(-1/2<x<1/2\).
(b) The inner derivative is \(-3\), so \(y'=3/\sqrt{1-(1-3x)^2}\).
(c) \(y'=2x/(1+x^4)\), defined for all real \(x\).
(d) By the product rule, \(y'=\arcsin x+x/\sqrt{1-x^2}\).
(e) The point is \((1,\pi/4)\) and the slope is \(1/(1+1^2)=1/2\), so \(y-\frac{\pi}{4}=\frac12(x-1)\).
(f) \(y'=2x/\sqrt{1-(x^2-2)^2}\). The function domain is \([-\sqrt3,-1]\cup[1,\sqrt3]\); the derivative domain is \((-\sqrt3,-1)\cup(1,\sqrt3)\).
(g) From \(\sin y=x\), obtain \(\cos y\,y'=1\). On the principal range, \(\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2}\), so \(y'=1/\sqrt{1-x^2}\) for \(|x|<1\).
(h) \(y'=3/(|3x|\sqrt{9x^2-1})=1/(|x|\sqrt{9x^2-1})\), which exists for \(|x|>1/3\).