Why Rational Inputs Require Care
A rational expression can make an inverse-trigonometric derivative look complicated, but the calculation follows a reliable pattern: identify the inner function, differentiate it with the quotient rule, and then apply the chain rule. The main extra responsibility is tracking the domain. A denominator can exclude inputs, and arcsine and arccosine require their inputs to lie between \(-1\) and \(1\).
These checks matter because a formula for a derivative applies only where the composite is defined and the relevant inverse-trigonometric derivative exists. A correct algebraic expression is not enough if it is used outside those conditions.
The Derivative Formulas and Their Conditions
Let \(u(x)\) be differentiable at \(x\). The inverse-trigonometric derivative rules give the following formulas. For arcsine and arccosine, the input must be strictly between \(-1\) and \(1\) at the point where the derivative is taken. Arctangent has no such range restriction.
If the inner function is rational, write it as \(u(x)=\frac{p(x)}{q(x)}\), where \(q(x)\ne 0\). At points where \(p\) and \(q\) are differentiable and \(q(x)\ne0\), the quotient rule gives
Substituting this expression into the appropriate inverse-trigonometric rule produces the derivative. The quotient rule does not remove the need to check the composite’s domain: the original rational expression must still have a nonzero denominator.
A Reliable Order of Work
Write the rational expression as \(u(x)\), so the outer function and inner function are easy to distinguish.
Exclude zeros of the rational denominator. For arcsine and arccosine, also require \(-1\le u(x)\le1\) for the function to be defined.
Use the quotient rule, keeping the numerator in the order \(p'q-pq'\).
For arcsine and arccosine, the displayed derivative formulas require the strict condition \(-1<u(x)<1\). For arctangent, any real \(u(x)\) is allowed.
Substitute a convenient point into the inner function, its derivative, and the final formula to confirm signs and values.
Why the Procedure Works
The chain rule explains why the derivative of the outer function is multiplied by \(u'(x)\). If \(F\) is an inverse-trigonometric function and \(y=F(u(x))\), then differentiating the composite gives \(y'=F'(u(x))u'(x)\). The rational input affects the calculation through its derivative \(u'(x)\), which is found with the quotient rule.
For example, when \(F=\arcsin\), the derivative of the outer function at an input \(z\) in \((-1,1)\) is \(F'(z)=1/\sqrt{1-z^2}\). Set \(z=u(x)\), then multiply by \(u'(x)\):
The condition \(-1<u(x)<1\) is essential in this argument: it ensures that the arcsine derivative formula is defined at the inner value. The arccosine and arctangent rules follow by the same chain-rule reasoning, using their respective outer derivatives.
Worked Examples
Worked Example: Arcsine of a Rational Function
Differentiate \(f(x)=\arcsin\left(\frac{x}{x^2+1}\right)\) and evaluate the derivative at \(x=1\).
Step 1: Check the domain. The denominator \(x^2+1\) is positive for every real \(x\), so the rational input is defined everywhere. Also, \((|x|-1)^2\ge0\), so \(x^2+1\ge2|x|\). Therefore \(\left|\frac{x}{x^2+1}\right|\le\frac12\), which lies strictly inside \((-1,1)\). Thus the composite is defined and the arcsine derivative rule applies for every real \(x\).
Step 2: Differentiate the inner function. Let \(u(x)=\frac{x}{x^2+1}\). The numerator has derivative \(1\), and the denominator has derivative \(2x\):
Step 3: Apply the arcsine chain rule.
Step 4: Evaluate at \(x=1\). The inner value is \(u(1)=\frac{1}{2}\), and the inner derivative is \(u'(1)=\frac{1-1}{(1+1)^2}=0\). Hence
Check: The denominator in the derivative formula is positive at \(x=1\), since \(\sqrt{1-(1/2)^2}=\sqrt{3}/2\). The zero derivative therefore comes from the inner derivative \(u'(1)=0\), as the chain rule predicts.
Worked Example: Arccosine and the Composite Domain
Find the domain of \(g(x)=\arccos\left(\frac{x-1}{x+1}\right)\), then differentiate it on the interval where the derivative formula applies.
Step 1: Find where the rational expression is defined. Its denominator requires \(x\ne-1\). Arccosine is defined when its input is between \(-1\) and \(1\), inclusive. Since \(x\ne-1\), we can compare squares with the positive denominator \((x+1)^2\):
Thus the composite’s domain is \([0,\infty)\). At the endpoint \(x=0\), its inner value is \(-1\); the function is defined there, but the standard arccosine derivative formula requires a strict inequality. We therefore use that formula for \(x>0\).
Step 2: Differentiate the rational input. Let \(u(x)=\frac{x-1}{x+1}\). Then \(u'(x)=1\) and the denominator’s derivative is \(1\), so
Step 3: Apply the arccosine rule. For \(x>0\), the inner value is strictly between \(-1\) and \(1\), giving
Step 4: Evaluate at \(x=3\). Here \(u(3)=\frac{2}{4}=\frac12\), and \(u'(3)=\frac{2}{16}=\frac18\). Therefore
Check: The direct derivative formula gives \(-2/(16\cdot\sqrt{3/4})=-2/(16\cdot\sqrt{3}/2)=-1/(4\sqrt{3})\), agreeing with the chain-rule evaluation. The negative sign comes from the arccosine derivative rule.
Worked Example: Arctangent of a Rational Function
Differentiate \(h(x)=\arctan\left(\frac{2x+1}{x-1}\right)\) and find \(h'(2)\).
Step 1: Check the domain. Arctangent accepts every real input, so the only restriction comes from the rational denominator: \(x\ne1\).
Step 2: Differentiate the inner function. Set \(u(x)=\frac{2x+1}{x-1}\). The numerator derivative is \(2\), and the denominator derivative is \(1\):
Step 3: Apply the arctangent rule.
For \(x\ne1\), combine the terms in the denominator:
Consequently,
Step 4: Evaluate at \(x=2\). The input is \(u(2)=\frac{5}{1}=5\), and \(u'(2)=-3\). Thus
Check: Substituting \(x=2\) into the simplified derivative gives \(-3/(5\cdot4+2\cdot2+2)=-3/26\), the same result.
Common Mistake: Losing the Outer Sign
A frequent error is to use the arcsine derivative rule for arccosine. For the function in the second example, a mistaken calculation might write
This has the wrong sign: the arccosine derivative is negative. The quotient-rule derivative \(u'(x)=2/(x+1)^2\) is positive, but it is multiplied by the negative outer derivative. The corrected expression is
Another important check is whether the inverse-trigonometric input is in range. For arcsine and arccosine, the composite function can be defined at an endpoint where \(u(x)=1\) or \(u(x)=-1\), while the displayed derivative formulas still do not apply there. Do not treat “function is defined” and “this derivative formula applies” as the same claim.
What to Remember
For a rational input, the computation has two layers: the quotient rule finds the rate of change of the inner expression, and the chain rule combines that rate with the inverse-trigonometric derivative. Domain checks come before the final formula: the rational denominator must be nonzero, and arcsine or arccosine inputs must be in \([-1,1]\) for the function to exist. Their derivative formulas require the stricter condition \(-1<u(x)<1\).
Check Your Understanding
For each question, identify the domain conditions before applying a derivative rule.
- For \(f(x)=\arcsin\left(\frac{x+1}{x^2+2}\right)\), find the inner derivative and state why the arcsine derivative formula is applicable wherever the rational expression is defined.
- For \(g(x)=\arccos\left(\frac{2x}{x+1}\right)\), what conditions must be checked before differentiating?
- Differentiate \(h(x)=\arctan\left(\frac{x}{x-2}\right)\), stating the excluded input.
- Explain why \(u(x)=1\) is allowed in the domain of \(\arcsin(u(x))\) but does not satisfy the condition in the usual arcsine derivative formula.
- For \(r(x)=\arctan\left(\frac{3x-1}{x+2}\right)\), find \(r'(0)\) and verify it by substituting \(x=0\) into your derivative expression.
