Why Radical Simplification Needs Care
Derivatives of arcsine and arccosine compositions contain a radical in the denominator. The expression under that radical often factors neatly, but simplifying the square root can change the sign information in the derivative if the domain is ignored.
The key issue is that a square root is always nonnegative. In particular, \(\sqrt{h(x)^2}\) is not always \(h(x)\); it is \(|h(x)|\). The absolute value may disappear only after the relevant domain tells us the sign of \(h(x)\).
The Derivative Formulas and Their Domains
For a differentiable function \(g\), the composition formulas are
The strict inequality matters: these formulas apply where the input is inside the interval \((-1,1)\). At an endpoint, where \(g(x)=1\) or \(g(x)=-1\), the denominator is zero, so the formulas do not give a finite derivative there. The composite function may still be defined at such a point, but differentiability must be checked separately.
A common situation is a rational input \(g(x)=p(x)/q(x)\), with \(q(x)\ne 0\). Then
Factoring the numerator may reveal a square. But taking the square root of the denominator gives \(\sqrt{q(x)^2}=|q(x)|\), not automatically \(q(x)\).
Why Absolute Values Appear
For every real number \(a\), \(\sqrt{a^2}\) is the nonnegative number whose square is \(a^2\). If \(a\ge 0\), that number is \(a\); if \(a<0\), it is \(-a\). These two cases are exactly the definition of absolute value:
More generally, if a radicand factors as \(h(x)^2r(x)\), with \(r(x)\ge 0\), then
To remove the absolute value, determine the sign of \(h(x)\) on the interval being considered. If \(h(x)>0\), then \(|h(x)|=h(x)\); if \(h(x)<0\), then \(|h(x)|=-h(x)\). When the sign changes, the simplified derivative may need a piecewise form.
Worked Example: A Difference of Squares with a Sign Change
Worked Example: Simplifying an Arcsine Derivative on Two Intervals
Differentiate \(y=\arcsin\left(\frac{2x}{1+x^2}\right)\), and state where the derivative exists.
Let \(g(x)=\frac{2x}{1+x^2}\). Its derivative is
Factor the radicand:
Because \(1+x^2>0\), its absolute value is \(1+x^2\). Therefore,
The input satisfies \(|g(x)|<1\) except at \(x=-1\) and \(x=1\), where it equals \(-1\) and \(1\), respectively. Apply the arcsine derivative formula away from those endpoints:
For \(-1<x<1\), \(|x^2-1|=1-x^2\), so \(y'=\frac{2}{1+x^2}\). For \(x<-1\) or \(x>1\), \(|x^2-1|=x^2-1\), while \(1-x^2=-(x^2-1)\), so \(y'=-\frac{2}{1+x^2}\). Thus
Check: On \((-1,1)\), the identity \(\arcsin\left(\frac{2x}{1+x^2}\right)=2\arctan(x)\) gives derivative \(\frac{2}{1+x^2}\). For \(x>1\), the composition equals \(2\arctan(1/x)\), whose derivative is \(-\frac{2}{1+x^2}\); for \(x<-1\), it equals \(-2\arctan(1/|x|)=2\arctan(1/x)\), again with derivative \(-\frac{2}{1+x^2}\). At \(x=\pm1\), the derivative does not exist: the composition has a corner at each point.
Worked Example: A Factor of \(x^2\)
Worked Example: Simplifying an Arccosine Derivative
Differentiate \(y=\arccos\left(\frac{x^2-1}{x^2+1}\right)\) wherever the derivative exists.
Set \(g(x)=\frac{x^2-1}{x^2+1}\). The quotient rule gives
Now factor the radicand:
Since \(x^2+1>0\),
The input lies between \(-1\) and \(1\) for every finite \(x\ne 0\). At \(x=0\), it equals \(-1\), so the formula for the derivative does not apply there. For \(x\ne 0\), the arccosine formula gives
If \(x>0\), then \(x/|x|=1\), giving \(y'=-\frac{2}{1+x^2}\). If \(x<0\), then \(x/|x|=-1\), giving \(y'=\frac{2}{1+x^2}\). Therefore,
Check: For \(x>0\), \(y=2\arctan(1/x)\), whose derivative is \(-\frac{2}{1+x^2}\). For \(x<0\), \(y=2\arctan(-1/x)\), whose derivative is \(\frac{2}{1+x^2}\). At \(x=0\), the composition has a corner, so it is not differentiable there.
Worked Example: Using the Domain to Remove an Absolute Value
Worked Example: Simplifying a Rational Arcsine Input
Differentiate \(y=\arcsin\left(\frac{x-1}{x+1}\right)\) on its domain, and simplify the result.
The input is defined when \(x\ne -1\). For the arcsine derivative formula, require
Squaring both sides, which is valid because both sides are nonnegative, gives \((x-1)^2<(x+1)^2\). Expanding and simplifying, \(x^2-2x+1<x^2+2x+1\), so \(x>0\). Thus the derivative formula applies on \(x>0\).
Differentiate the inner function:
Factor the radicand:
For \(x>0\), the denominator \(x+1\) is positive. Hence
Substitute into the arcsine derivative formula:
Check: The simplified derivative is positive for every \(x>0\), as is the formula’s numerator \(g'(x)\) and denominator \(\sqrt{1-g(x)^2}\). Substituting the displayed radical and \(g'(x)\) back into their quotient gives \(\frac{2}{(x+1)^2}\cdot\frac{x+1}{2\sqrt{x}}=\frac{1}{(x+1)\sqrt{x}}\), as claimed. At \(x=0\), the input is \(-1\), so the derivative formula is not applicable.
A Common Mistake: Dropping the Absolute Value
In the second example, a tempting but incorrect step is
This is wrong for \(x<0\), because the left side is nonnegative while the proposed right side is negative. The correct simplification is
Dropping \(|x|\) would change the sign of the resulting derivative for negative \(x\). The correction is not to guess the sign, but to retain the absolute value until the domain is split into intervals where the sign is known.
A Reliable Simplification Process
When a derivative contains \(\sqrt{1-g(x)^2}\), use this sequence. It keeps the algebra, the domain, and the sign decisions distinct.
For arcsine and arccosine, first identify where \(|g(x)|<1\). Also exclude any point where \(g\) is undefined.
Rewrite \(1-g(x)^2\), combine terms, and look for difference-of-squares factors or perfect squares.
Replace each square root of a square by an absolute value, and use \(|q(x)|\) for a square-rooted denominator \(q(x)^2\).
Determine the sign of each expression inside an absolute value on the relevant interval. State separate cases if its sign changes.
Substitute the simplified radical back into the derivative formula, and confirm that the sign and domain agree with the original expression.
Why the Domain Check Matters
Factoring is not merely a way to make an expression shorter. It can reveal where the composition is defined, where its derivative formula applies, and where a sign change forces separate answers. For example, a factor \(x^2\) under a square root becomes \(|x|\), which behaves differently on \(x<0\) and \(x>0\). Similarly, a denominator such as \(x+1\) can be replaced by its absolute value only after deciding whether the domain lies to the left or right of \(-1\).
Keep endpoint points separate from interior points. A composite arcsine or arccosine can be defined when its input equals \(1\) or \(-1\), but the derivative formulas above require a strictly interior input. A zero denominator in the formula is a warning to check differentiability directly, not permission to simplify by cancellation.
Check Your Understanding
For each problem, pay attention to both the radical simplification and the interval where the derivative formula applies.
- For \(y=\arcsin\left(\frac{x^2-4}{x^2+4}\right)\), factor \(1-g(x)^2\) and state the absolute-value expression for its square root.
- Why is \(\sqrt{(x-3)^2}=|x-3|\), rather than always \(x-3\)?
- For \(y=\arccos\left(\frac{x}{\sqrt{x^2+1}}\right)\), identify where the input is strictly between \(-1\) and \(1\).
- When can \(|x+2|\) be replaced by \(x+2\)? When must it be replaced by \(-(x+2)\)?
- What must be checked separately when an arcsine or arccosine input equals \(1\) or \(-1\)?
