Introduction
Expressions involving inverse trigonometric functions can combine several layers of differentiation: an inverse function may have a polynomial input, appear in a product, or sit in the numerator of a quotient. The main challenge is not a new differentiation rule. It is keeping the rules, domains, and signs organized at every layer.
This workshop develops a reliable way to handle such expressions. At each stage, identify the outer operation, differentiate its pieces, and check that the expression is defined and differentiable at the point under consideration. Particular care is needed for arcsecant, whose derivative contains an absolute value.
Derivative rules and their conditions
Write \(\arcsin u\), \(\arccos u\), and so on for the corresponding inverse trigonometric functions applied to an input \(u=u(x)\). The chain rule multiplies the derivative of the outer inverse function by \(u'(x)\). The formulas below apply wherever the input is in the relevant domain and the displayed derivative exists.
The function values \(\arcsin u\) and \(\arccos u\) are defined for \(|u|\leq 1\), but their derivative formulas have a zero denominator at \(u=\pm1\). Likewise, arcsecant and arccosecant are defined for \(|u|\geq1\), while their displayed derivatives require \(|u|>1\). A composition may be defined at an endpoint even though its derivative formula does not apply there.
For the algebraic operations, if \(p\) and \(q\) are differentiable, then \((pq)'=p'q+pq'\). If \(q\ne0\), then \((p/q)'=(p'q-pq')/q^2\). These rules are used together with the chain rule, not instead of it.
Why the rules can be combined
The chain rule explains why a nested input contributes an additional factor. Suppose \(F\) is differentiable at \(u(x)\), and \(u\) is differentiable at \(x\). For a nonzero increment \(h\), the difference quotient can be written as
whenever \(u(x+h)\ne u(x)\). As \(h\) tends to zero, the first factor tends to \(F'(u(x))\), and the second tends to \(u'(x)\). If some increments make the first quotient unavailable, the same conclusion follows by the differentiability expansion \(F(u(x)+k)=F(u(x))+F'(u(x))k+o(k)\), with \(k=u(x+h)-u(x)\). Thus the derivative is \(F'(u(x))u'(x)\).
For a product, the product rule accounts for the two ways a small change can affect the factors: \(p\) changes while \(q\) is held at its current value, and \(q\) changes while \(p\) is held at its current value. The quotient rule follows by differentiating \(p/q=pq^{-1}\), where \(q\ne0\), and applying the chain and product rules. In a mixed expression, these rules are applied successively to the actual outer structure of the expression.
Check inverse-function inputs, square roots, and any denominator before using a derivative formula.
Decide whether the expression is a product, quotient, or a single composition.
Use the relevant inverse trigonometric rule and multiply by the derivative of its input.
Retain the negative signs and absolute values required by the formulas, then test the result at a valid point if possible.
Worked Examples
Worked Example: A product with a nested arcsine
Differentiate \(y=(x^2+1)\arcsin(x/2)\), and evaluate \(y'\) at \(x=1\).
The arcsine input must satisfy \(|x/2|\leq1\), so the function is defined for \(-2\leq x\leq2\). The derivative formula for arcsine applies when \(|x/2|<1\), giving the open interval \(-2<x<2\).
Set \(p=x^2+1\) and \(q=\arcsin(x/2)\). Then \(p'=2x\). By the chain rule, with inner input \(u=x/2\) and \(u'=1/2\),
Now apply the product rule:
At \(x=1\), the inverse-function input is \(1/2\), which is strictly between \(-1\) and \(1\). Also, \(\arcsin(1/2)=\pi/6\) and \(\sqrt{4-1}=\sqrt3\). Therefore
Both terms are positive, so the derivative at this point is positive. Checking the product-rule pieces separately gives \(p'(1)=2\), \(q(1)=\pi/6\), \(p(1)=2\), and \(q'(1)=1/\sqrt3\); the sum \(p'(1)q(1)+p(1)q'(1)\) reproduces the result.
Worked Example: A quotient with an arccosine input
Differentiate \(y=\frac{\arccos(x/3)}{1+x^2}\), and find \(y'(1)\).
For \(|x|\leq3\), \(|x/3|\leq1\), and the denominator \(1+x^2\) is positive for every real \(x\). The derivative of the arccosine composition uses the strict condition \(|x/3|<1\), so the formula below applies on \(-3<x<3\).
Let \(p=\arccos(x/3)\) and \(q=1+x^2\). The chain rule gives
Using the quotient rule,
At \(x=1\), the denominator is \(2^2=4\), the square root is \(\sqrt8=2\sqrt2\), and \(\arccos(1/3)\) is defined. Thus
The result is negative: \(\arccos(1/3)>0\), and both terms in the final expression are negative. To check the quotient-rule substitution, \(p'(1)=-1/(2\sqrt2)\), \(q(1)=2\), \(p(1)=\arccos(1/3)\), and \(q'(1)=2\); the numerator \(p'q-pq'\) is exactly \(-1/\sqrt2-2\arccos(1/3)\).
Worked Example: A quotient with two nested inverse functions
Differentiate \(y=\frac{\arctan(2x)}{2+\arcsin(x/2)}\), and evaluate the derivative at \(x=0\).
The arcsine input requires \(|x/2|\leq1\), so the expression is defined for \(-2\leq x\leq2\). On that interval, \(-\pi/2\leq\arcsin(x/2)\leq\pi/2\). Consequently, the denominator is at least \(2-\pi/2>0\); it never vanishes. The derivative formulas for arcsine apply in the interior \(-2<x<2\).
Write \(p=\arctan(2x)\) and \(q=2+\arcsin(x/2)\). The inner derivative of \(2x\) is \(2\), while that of \(x/2\) is \(1/2\), so
Applying the quotient rule gives
At \(x=0\), \(p(0)=\arctan(0)=0\), \(p'(0)=2\), \(q(0)=2\), and \(q'(0)=1/2\). Hence
This is also a direct check using the quotient rule at zero: the numerator of the derivative is \(p'(0)q(0)-p(0)q'(0)=4\), and the denominator is \(q(0)^2=4\). The ratio is \(1\), as found.
Worked Example: A product that requires the arcsecant absolute value
Differentiate \(y=\operatorname{arccot}(x^2-1)\operatorname{arcsec}(2-x)\), and evaluate \(y'(0)\).
The arccotangent input is real for every \(x\). The arcsecant input must satisfy \(|2-x|\geq1\), which means \(x\leq1\) or \(x\geq3\). Its derivative formula requires strict inequality, so the product is differentiable on \(x<1\) and \(x>3\); the point \(x=3\) is in the function's domain but not covered by that derivative formula.
Set \(r=\operatorname{arccot}(x^2-1)\) and \(s=\operatorname{arcsec}(2-x)\). The chain rule gives
Thus, on \(x<1\) or \(x>3\), the product rule yields
At \(x=0\), the arcsecant input is \(2\), so its absolute value is \(2\), and its square-root factor is \(\sqrt3\). Also \(\operatorname{arccot}(-1)=3\pi/4\) for the principal arccotangent range \((0,\pi)\). Since \(r'(0)=0\) and \(s'(0)=-1/(2\sqrt3)\),
The sign is negative because the arccotangent value is positive while the arcsecant factor is decreasing at \(x=0\). This agrees with the product-rule expression: its first term vanishes at zero and its second term is negative.
A common sign error
The absolute value in the arcsecant derivative cannot be discarded, even when the input is negative. For example, take \(y=\operatorname{arcsec}(-2x)\) at \(x=1\). The input is \(-2\), which is in the arcsecant derivative domain, and the inner derivative is \(-2\). The correct calculation is
If the absolute value were incorrectly omitted, the denominator would be \((-2)\sqrt3\), producing \(+1/\sqrt3\), the wrong sign. A useful check is that the denominator \(|u|\sqrt{u^2-1}\) is positive wherever \(|u|>1\); the sign of the derivative therefore comes from \(u'\) alone for arcsecant, and from \(-u'\) for arccosecant.
Why domain checks belong in the calculation
A formula can look algebraically correct and still be used where it does not apply. For example, an arcsine composition can be defined when its input equals \(1\), but its derivative formula contains \(\sqrt{1-u^2}=0\) there. Similarly, a quotient requires a nonzero denominator, and arcsecant's derivative formula requires its input to be strictly outside \([-1,1]\). Record these restrictions before simplifying a derivative.
In practice, keep the expression in layers until each layer has been differentiated. For a quotient of inverse-function compositions, first find the derivatives of the numerator and denominator separately, including each inner derivative. Then assemble the quotient rule. This reduces the chance of dropping a chain-rule factor or moving a negative sign into the wrong term.
Check Your Understanding
For each question, identify the domain restriction before applying a derivative rule.
- For \(f(x)=(x+1)\arccos(x/2)\), what is the interval on which the derivative formula for the arcsine or arccosine factor applies?
- Differentiate \(\frac{\arctan(x^2)}{1+x}\), stating the restriction imposed by the denominator.
- For \(g(x)=\operatorname{arcsec}(1-x)\), where is the function defined, and where does the standard derivative formula apply?
- In differentiating \(\operatorname{arccsc}(3x)\), which sign appears in front of the chain-rule factor?
- Explain why replacing \(|u|\) by \(u\) in the arcsecant derivative can give an incorrect sign when \(u< -1\).
