From Arcsine to an Arcsine Composition
The derivative of an arcsine function becomes a chain-rule problem when its input is itself a function of \(x\). The key is to track two things at once: how quickly the inner function changes, and whether its values lie where the arcsine derivative formula applies.
The function \(\arcsin(u)\) is defined for \(-1\leq u\leq 1\), and its values lie in the interval \([-\pi/2,\pi/2]\). Its derivative formula, \(\frac{1}{\sqrt{1-u^2}}\), applies when \(-1<u<1\). The endpoints are excluded because the denominator is zero there.
The composition itself may be defined at points where \(g(x)=1\) or \(g(x)=-1\). That does not make the derivative formula valid there: its denominator would be zero. At such a point, the derivative of the composition needs separate analysis.
Why the Formula Works
The chain rule says that differentiating an outer function applied to an inner function multiplies the outer derivative, evaluated at the inner function, by the derivative of that inner function. Here, the arcsine derivative contributes the denominator, and \(g'(x)\) accounts for the inner function’s rate of change.
The same result can be justified directly using the inverse relationship between sine and arcsine. Fix a point \(x\) where \(g\) is differentiable and \(-1<g(x)<1\), and set \(y=\arcsin(g(x))\). Then \(y\) lies strictly between \(-\pi/2\) and \(\pi/2\), so \(\cos y>0\).
Since \(y=\arcsin(g(x))\), we have \(\sin y=g(x)\).
By the chain rule, \(\cos y\,\frac{dy}{dx}=g'(x)\).
Because \(\cos y>0\), the identity \(\sin^2 y+\cos^2 y=1\) gives \(\cos y=\sqrt{1-\sin^2 y}=\sqrt{1-(g(x))^2}\).
The square root is positive because \(-1<g(x)<1\), so dividing by it gives \(\frac{dy}{dx}=\frac{g'(x)}{\sqrt{1-(g(x))^2}}\).
This argument also shows why the square root is positive rather than negative: the principal range of arcsine ensures that its corresponding angle has positive cosine.
Finding the Required Input Restrictions
Before differentiating, identify where the composition is defined. For \(\arcsin(g(x))\), that means requiring \(-1\leq g(x)\leq 1\). To use the derivative formula, impose the stricter condition \(-1<g(x)<1\), and also require \(g\) to be differentiable at the point under consideration.
When these conditions hold throughout an interval, the formula applies throughout that interval. In practice, it is often useful to solve the range condition first, then differentiate only on the resulting region. Checking restrictions after the calculation can hide points where the formula was never valid.
Worked Example: A Linear Inner Function
Differentiate \(f(x)=\arcsin(3x-1)\), and state where the derivative formula applies.
First, the composition is defined when its input is between \(-1\) and \(1\), inclusive:
Add \(1\) throughout, then divide by \(3\):
For the derivative formula, the input must be strictly between the endpoints, so \(0<x<\frac{2}{3}\). The inner function \(g(x)=3x-1\) has derivative \(g'(x)=3\). Therefore,
As a check, at \(x=\frac{1}{3}\) the inner value is \(3(\frac{1}{3})-1=0\), and the formula gives \(f'(\frac{1}{3})=3/\sqrt{1}=3\). In the inverse-relation check, \(f(\frac{1}{3})=\arcsin(0)=0\), so \(\cos(f(\frac{1}{3}))=1\); differentiating \(\sin(f(x))=3x-1\) gives \(\cos(f(x))f'(x)=3\), hence \(f'(\frac{1}{3})=3\), as required.
Polynomial Inputs and Simplification
For a polynomial inner function, apply the same rule: square the entire inner expression in the denominator, and multiply by its derivative in the numerator. Parentheses matter because the denominator contains \(1-(g(x))^2\), not \(1-g(x)\).
Worked Example: A Squared Input
Differentiate \(h(x)=\arcsin(x^2)\), and evaluate the derivative at \(x=\frac{1}{2}\).
The composition is defined when \(-1\leq x^2\leq 1\). Since \(x^2\geq 0\), this is equivalent to \(x^2\leq 1\), or \(-1\leq x\leq 1\). The derivative formula requires \(x^2<1\), so it applies for \(-1<x<1\). The inner derivative is \(2x\), giving
At \(x=\frac{1}{2}\), the numerator is \(2(\frac{1}{2})=1\), and the denominator is
Thus
For a check, \(h(\frac{1}{2})=\arcsin(\frac{1}{4})\), whose sine is \(\frac{1}{4}\) and whose cosine is \(\sqrt{15}/4\). Differentiating \(\sin(h(x))=x^2\) gives \(\cos(h(x))h'(x)=2x\). At \(x=\frac{1}{2}\), this is \((\sqrt{15}/4)h'(\frac{1}{2})=1\), so \(h'(\frac{1}{2})=4/\sqrt{15}\), matching the result.
Radical Inputs and Absolute Values
A radical inside the arcsine can make the restrictions and simplification more delicate. In particular, simplifying a square root such as \(\sqrt{x^2}\) requires writing \(|x|\), not \(x\). That distinction can affect the sign of a derivative.
Worked Example: A Square-Root Input
Differentiate \(p(x)=\arcsin\!\left(\sqrt{1-x^2}\right)\) using the composition formula, and find \(p'(\frac{3}{5})\).
The radical requires \(1-x^2\geq 0\), so \(-1\leq x\leq 1\). On this interval, \(\sqrt{1-x^2}\) lies between \(0\) and \(1\), so the arcsine composition is defined. For the derivative formula, its input must be strictly less than \(1\), which excludes \(x=0\). The inner function \(g(x)=\sqrt{1-x^2}\) is differentiable for \(-1<x<1\), so the formula applies on \((-1,0)\) and \((0,1)\).
Differentiate the inner function:
Substitute into the arcsine formula. In the denominator, \(1-(g(x))^2=1-(1-x^2)=x^2\), so \(\sqrt{x^2}=|x|\):
At \(x=\frac{3}{5}\), \(|x|=\frac{3}{5}\) and \(\sqrt{1-x^2}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\). Therefore,
To check, \(g(\frac{3}{5})=\frac{4}{5}\), so \(\cos(p(\frac{3}{5}))=\sqrt{1-(\frac{4}{5})^2}=\frac{3}{5}\). Also \(g'(\frac{3}{5})=-(\frac{3}{5})/(\frac{4}{5})=-\frac{3}{4}\). Differentiating \(\sin(p(x))=g(x)\) gives \(\cos(p(x))p'(x)=g'(x)\), so \(p'(\frac{3}{5})=(-\frac{3}{4})/(\frac{3}{5})=-\frac{5}{4}\).
Common Mistakes and Endpoint Caution
A frequent error is to differentiate the arcsine but forget to multiply by the derivative of its input. For example, the incorrect calculation \(\frac{d}{dx}\arcsin(3x)=\frac{1}{\sqrt{1-9x^2}}\) treats \(3x\) as if its derivative were \(1\).
Another mistake is to use the formula at an endpoint merely because the composition is defined there. For instance, \(\arcsin(\sqrt{1-x^2})\) is defined at \(x=0\), but its arcsine input equals \(1\), so the denominator in the formula is zero. In fact, on \([-1,1]\) this composition equals \(\pi/2-|x|\), which is not differentiable at \(0\). The endpoint formula does not establish a derivative there.
More generally, when \(g(x)=1\) or \(g(x)=-1\), do not substitute into the formula or assume the composition has no derivative. The composition may fail to be differentiable, as in the example, or the behavior may require a separate limit argument. The formula proved here makes a precise claim only when \(g\) is differentiable and its value is strictly inside \((-1,1)\).
Check Your Understanding
For each question, identify where the derivative formula applies before simplifying or evaluating.
- Find the derivative of \(\arcsin(2x+1)\), and state the interval on which the formula is valid.
- Find the derivative of \(\arcsin(1-x^2)\), and state the restrictions on \(x\).
- For \(q(x)=\arcsin(x^3)\), find \(q'(x)\) and evaluate it at \(x=\frac{1}{2}\).
- Why is \(\sqrt{x^2}\) equal to \(|x|\), and how can replacing it with \(x\) give a wrong derivative?
- Does the arcsine composition derivative formula apply at a point where the inner value is \(1\)? Explain why or why not.
