Quadratic Inputs and the Chain Rule
A quadratic inside an inverse trigonometric function makes the derivative depend on two things: the derivative of the quadratic and the derivative rule for the outer function. The Chain Rule combines them. A separate domain check tells us where the original composition exists and where its derivative formula applies.
For a real quadratic \(q(x)=ax^2+bx+c\), where \(a\), \(b\), and \(c\) are constants, the basic formulas are:
The first composition is defined for every real value of \(q(x)\). The second is defined when \(-1\le q(x)\le 1\), but the displayed derivative formula requires the stricter condition \(-1<q(x)<1\). At an endpoint, the function can be defined even though the derivative formula does not apply.
Why the Formulas Work
The Chain Rule says that if \(F\) is differentiable at \(q(x)\) and \(q\) is differentiable at \(x\), then the derivative of \(F(q(x))\) is \(F'(q(x))q'(x)\). The inverse trigonometric derivative rules give the outer derivatives. Substituting the inner function \(q(x)\) yields the formulas above.
For the quadratic \(q(x)=ax^2+bx+c\), its derivative is \(q'(x)=2ax+b\). The main task is therefore to substitute this derivative and the original quadratic accurately, then simplify without changing the expression’s domain.
Worked Examples
Worked Example: Differentiate \(\arctan(x^2)\)
Let \(y=\arctan(x^2)\). The inner function is \(q(x)=x^2\), so \(q'(x)=2x\). Since arctangent is defined for every real input, there is no restriction on \(x\). Applying the Chain Rule gives
The denominator \(1+x^4\) is positive for every real \(x\), so the derivative exists for all real \(x\). To check, differentiate the original outer function using the arctangent rule: the outer derivative is \(1/(1+(x^2)^2)\), and multiplying by the inner derivative \(2x\) gives \(2x/(1+x^4)\), as obtained.
Worked Example: Differentiate \(\arctan(3x^2-2x+1)\)
Set \(q(x)=3x^2-2x+1\). Its derivative is \(q'(x)=6x-2\). The arctangent input may be any real number, so the composition is defined for all real \(x\). By the Chain Rule,
This is already a valid simplified form; expanding the square is unnecessary. Its denominator is at least \(1\), so it never vanishes. As a check, the derivative of the inner quadratic is \(6x-2\), and the arctangent rule contributes the denominator \(1+q(x)^2\). Substituting \(q(x)=3x^2-2x+1\) reproduces the result.
Worked Example: Find the domain and derivative of \(\arcsin(x^2/2)\)
The arcsine input is \(q(x)=x^2/2\). The composition is defined when its input lies in \([-1,1]\). Since \(x^2/2\ge 0\), the condition becomes \(x^2/2\le 1\), or \(x^2\le 2\). Thus its domain is \([-\sqrt{2},\sqrt{2}]\).
For the derivative formula, the input must lie strictly between \(-1\) and \(1\). Here that means \(x^2/2<1\), so \(-\sqrt{2}<x<\sqrt{2}\). The inner derivative is \(q'(x)=x\). Therefore, on that open interval,
The radical is positive on the stated open interval. At \(x=\pm\sqrt{2}\), the composition is defined, but the derivative formula has a zero denominator, so it does not give a finite derivative there. To check the calculation in the interior, \(q'(x)=x\) and \(1-q(x)^2=1-x^4/4\); substituting these into the arcsine rule gives exactly the displayed expression.
A Quadratic-Derived Input and an Absolute Value
Sometimes the input is a rational expression built from quadratic terms. For example, \(2x/(1+x^2)\) is defined for every real \(x\), since \(1+x^2>0\). It also lies in the arcsine domain: the inequality \(2|x|\le 1+x^2\) follows from \((|x|-1)^2\ge 0\). Equality holds at \(x=\pm1\). Thus the composition below is defined for all real \(x\), including those two points.
Worked Example: Differentiate \(\arcsin\left(\frac{2x}{1+x^2}\right)\) where possible
Write \(u(x)=2x/(1+x^2)\). Using the quotient rule,
For \(|u(x)|<1\), the arcsine derivative rule and Chain Rule give
To simplify, calculate the expression under the radical:
Since \(1+x^2>0\), its square root is
The absolute value is essential: in general, \(\sqrt{z^2}=|z|\), not \(z\). For \(x\ne -1,1\), substitution and cancellation now give
where \(\operatorname{sgn}(r)=1\) for \(r>0\) and \(\operatorname{sgn}(r)=-1\) for \(r<0\). This formula applies for \(x\ne\pm1\), the points where \(|u(x)|<1\).
At \(x=1\) and \(x=-1\), the function is defined but is not differentiable. To verify this, use \(u(x)=\sin(2\arctan x)\). Near \(x=1\), the composition equals \(2\arctan x\) for \(x<1\), and \(\pi-2\arctan x\) for \(x>1\). The one-sided derivatives at \(1\) are \(1\) and \(-1\), so there is no derivative there. Near \(x=-1\), the corresponding expressions are \(-\pi-2\arctan x\) for \(x<-1\) and \(2\arctan x\) for \(x>-1\). Their one-sided derivatives at \(-1\) are \(-1\) and \(1\), so there is no derivative there either.
As a check away from the endpoints, the unsimplified Chain Rule expression has numerator \(2(1-x^2)\) and radical \(|1-x^2|/(1+x^2)\). Cancelling the nonzero factor \(1-x^2\) leaves \(2\operatorname{sgn}(1-x^2)/(1+x^2)\), matching the simplified derivative.
Common Mistakes and a Reliable Check
A frequent error is to replace \(\sqrt{(1-x^2)^2}\) by \(1-x^2\). That replacement is false when \(|x|>1\), because \(1-x^2<0\) there while a square root cannot be negative. The correct equality is
Another error is to state a derivative at a point just because the original composition is defined there. The arcsine input may equal \(1\) or \(-1\), where the function exists, while the derivative formula has a zero denominator. The endpoint must be checked separately; in the rational-input example, the one-sided derivatives show that the composition is not differentiable at either endpoint.
For arctangent, every real input is allowed. For arcsine, require the input to lie in \([-1,1]\) for the function to be defined.
For a quadratic \(ax^2+bx+c\), the derivative is \(2ax+b\).
Multiply by the inner derivative, and for arcsine require the input to lie strictly between \(-1\) and \(1\).
When taking a square root of a square, retain the absolute value and check any points where cancellation would divide by zero.
These checks matter because a correct algebraic expression can still be an incomplete derivative answer if its domain is omitted. For a quadratic input, the Chain Rule is usually direct; the careful work is identifying the interval on which the derivative exists and preserving the signs during simplification.
Check Your Understanding
For each question, state any domain restriction as well as the derivative where it exists.
- Differentiate \(\arctan(4x^2+1)\).
- Find the domain of \(\arcsin(1-x^2)\), and state where the arcsine derivative formula applies.
- Why does \(\arcsin(x^2/2)\) have a defined value at \(x=\sqrt{2}\) even though the interior derivative formula does not apply there?
- Simplify \(\sqrt{(x^2-1)^2}\) correctly for all real \(x\).
- For \(\arcsin(2x/(1+x^2))\), which real inputs must be excluded from the Chain Rule derivative formula, and why?
