Why the Arcsine Derivative Needs a Branch
The sine function repeats its values, so it does not have an inverse on all real numbers. To define an inverse, we restrict sine to an interval where it is one-to-one. The choice of interval determines which angle arcsine returns, and that choice will also determine the sign in its derivative.
We will use implicit differentiation to derive the derivative, rather than treating the formula as something to memorize. The derivation will also show why the formula applies only between the endpoints of the arcsine domain.
The endpoints are included in the domain because \(\sin(-\pi/2)=-1\) and \(\sin(\pi/2)=1\). The derivative, however, will be finite only for inputs strictly between \(-1\) and \(1\).
Deriving the Formula by Implicit Differentiation
Set \(y=\arcsin(x)\). The defining equation is \(\sin y=x\). On the interior of the domain, \(y\) lies strictly between \(-\pi/2\) and \(\pi/2\). Differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\):
The chain rule gives \(\cos y\,\frac{dy}{dx}=1\). Because \(\cos y>0\) for \(-\pi/2<y<\pi/2\), we can divide by \(\cos y\). To express the result in terms of \(x\), use \(\sin y=x\) and the identity \(\sin^2 y+\cos^2 y=1\). The range of \(y\) ensures that \(\cos y\) is positive, so \(\cos y\) is the positive square root.
This argument establishes both the formula and its conditions. The inverse is differentiable at interior inputs because the slope of sine at the corresponding angle is \(\cos y\), which is nonzero there. At the endpoints, that slope is zero, so the division used in the derivation is not valid.
Let \(y=\arcsin(x)\), so \(\sin y=x\) and \(-\pi/2\le y\le\pi/2\).
The chain rule gives \(\cos y\,\frac{dy}{dx}=1\).
For interior inputs, \(\cos y>0\), so \(\cos y=\sqrt{1-\sin^2 y}\).
Since \(\sin y=x\), the result is \(\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\), for \(-1<x<1\).
Why the Square Root Is Positive
The identity \(\cos^2 y=1-\sin^2 y\) gives \(\cos y=\pm\sqrt{1-\sin^2 y}\) if considered without more information. Choosing a sign is not optional: the correct sign follows from the range used to define arcsine. On \([-\pi/2,\pi/2]\), cosine is nonnegative, and it is strictly positive in the interior. Therefore, for \(-1<x<1\),
The nonnegative square root is essential. Taking the negative square root would contradict the chosen range of arcsine and would produce a negative derivative, even though arcsine is increasing.
Worked Example: Derive the Derivative at an Arbitrary Interior Input
Let \(y=\arcsin(x)\), where \(-1<x<1\). Begin with the defining identity:
Differentiate both sides with respect to \(x\):
The range condition gives \(-\pi/2<y<\pi/2\), so \(\cos y>0\). Divide by \(\cos y\), then use \(\sin y=x\):
Check by substituting this result into the differentiated relation. Since \(\cos y=\sqrt{1-x^2}\),
This matches the derivative of the right-hand side \(x\), confirming the result for every interior input.
Using the Formula at Particular Inputs
To evaluate the derivative at an input, first check that the input lies in \((-1,1)\). Then substitute it into the formula and simplify the square root carefully. Because the square root is in the denominator, a smaller value of \(1-x^2\) produces a larger derivative.
Worked Example: Find the Derivative at \(x=-\sqrt{3}/2\)
The input is in the derivative's domain because \(-1<-\sqrt{3}/2<1\). Apply the formula:
Square the input and simplify the denominator:
Check the value by substituting it into the differentiated relation. At this input, \(\sin y=-\sqrt{3}/2\) and \(y\) is in the arcsine range, so \(\cos y=1/2\). Then \(\cos y\,\frac{dy}{dx}=(1/2)(2)=1\), as required.
The derivative is positive on its entire domain, consistent with arcsine being increasing. Its size depends on how close the input is to either endpoint.
Worked Example: Compare Slopes Near an Endpoint
Compare the derivative at \(x=3/5\) with the derivative at \(x=24/25\). Both inputs are in \((-1,1)\). At \(x=3/5\),
At \(x=24/25\),
Since \(25/7>5/4\), the derivative is larger at \(24/25\), which is closer to \(1\). Check each value in the derivative formula: \(1-(3/5)^2=16/25\), giving \(5/4\), and \(1-(24/25)^2=49/625\), giving \(25/7\). Both calculations use the positive square root, as required.
Behavior Near the Endpoints
As \(x\) approaches either endpoint from inside the domain, \(1-x^2\) approaches zero through positive values. Its square root therefore approaches zero through positive values, so its reciprocal grows without bound:
The one-sided directions matter: the arcsine domain ends at \(-1\) and \(1\), so the derivative formula approaches each endpoint only from within \((-1,1)\). The function has no finite derivative at either endpoint. In particular, substituting \(x=1\) or \(x=-1\) into the formula gives division by zero, not a derivative value.
The unbounded derivative also describes the steepening of the graph. At interior inputs the slope is finite, but as the input moves toward either endpoint, the slope becomes arbitrarily large. This is consistent with the fact that sine has slope zero at \(y=\pm\pi/2\): near those angles, a small change in sine corresponds to a comparatively larger change in the angle.
Common Mistake: Choosing the Wrong Sign
A frequent error is to use \(\cos y=-\sqrt{1-\sin^2y}\) simply because both signs can arise when solving an equation for cosine. For example, this would lead to the incorrect calculation
The correction is to use the range of arcsine before choosing the sign. Since \(-\pi/2<y<\pi/2\) for interior inputs, \(\cos y>0\), not \(\cos y<0\). Thus,
A related mistake is to state that this derivative formula applies on the closed interval \([-1,1]\). The function is defined at both endpoints, but the formula for a finite derivative is valid only on the open interval \((-1,1)\). Keeping the function's domain separate from the derivative's domain prevents that confusion.
Why the Derivation Matters
The arcsine derivative is a useful example of how an inverse-function derivative depends on the behavior of the original function. Implicit differentiation converts the relation \(\sin y=x\) into a slope equation. The selected inverse branch then determines the sign, while the zero slopes of sine at its extreme values explain why the inverse's derivative becomes unbounded near the endpoints.
For AP problems, keep three facts together: the arcsine input must be in \([-1,1]\); the finite derivative formula applies for \(-1<x<1\); and the square root in the denominator is positive on that open interval. These conditions make the calculation both accurate and meaningful.
Check Your Understanding
Use the defining relation and the derivative formula to answer these questions. State any domain restrictions you use.
- Why is \(\cos(\arcsin x)\) the positive square root of \(1-x^2\) for \(-1<x<1\)?
- Find \(\frac{d}{dx}\arcsin(x)\) at \(x=4/5\), showing the square-root simplification.
- Explain why the derivative formula does not give a finite value at \(x=-1\).
- What happens to the derivative as \(x\to1^-\), and why?
- Identify the error in claiming that \(\frac{d}{dx}\arcsin(x)=-1/\sqrt{1-x^2}\).
