Why Compositions Need the Chain Rule
A function such as \(\arcsin(x)\) applies an inverse trigonometric function directly to \(x\). A composition such as \(\arcsin(3x)\) applies it instead to an expression that changes with \(x\). Differentiating the composition must account for both layers: how the inverse trigonometric function changes as its input changes, and how that input changes as \(x\) changes.
The chain rule keeps those two rates of change connected. It says that the derivative of an outer function evaluated at an inner function is the outer derivative, evaluated at the inner function, multiplied by the derivative of the inner function.
Derivative Formulas and Their Conditions
The formulas below use the usual principal-value inverse trigonometric functions. In each row, \(u=g(x)\), where \(g\) is differentiable. The formula applies where the displayed denominator is defined and the inverse function is differentiable at \(u\). For the first three functions, the input must lie strictly between \(-1\) and \(1\); for arcsecant, its absolute value must be greater than \(1\).
| Composition | Derivative | Input condition |
|---|---|---|
| \(\arcsin(g(x))\) | \(\dfrac{g'(x)}{\sqrt{1-g(x)^2}}\) | \(|g(x)|<1\) |
| \(\arccos(g(x))\) | \(-\dfrac{g'(x)}{\sqrt{1-g(x)^2}}\) | \(|g(x)|<1\) |
| \(\arctan(g(x))\) | \(\dfrac{g'(x)}{1+g(x)^2}\) | All real \(g(x)\) |
| \(\arcsec(g(x))\) | \(\dfrac{g'(x)}{|g(x)|\sqrt{g(x)^2-1}}\) | \(|g(x)|>1\) |
For example, the arcsine rule has two distinct parts: \(1/\sqrt{1-u^2}\) is the derivative of \(\arcsin(u)\) with respect to \(u\), while \(g'(x)\) accounts for the change in \(u=g(x)\) as \(x\) changes. The same structure holds in every row. Notice that arcsecant has an absolute value in its denominator; its formula is not \(g'(x)/(g(x)\sqrt{g(x)^2-1})\) for all allowed inputs.
These conditions are local: to use a formula at a particular \(x\), the input must satisfy its condition there, and the composition must be defined and differentiable in a neighborhood of that point. For arcsine and arccosine, the endpoints \(u=\pm1\) are not included in the displayed derivative formulas because the denominator becomes zero. The arctangent formula has no corresponding restriction on real inputs.
Why the Chain Rule Gives These Results
Here is a justification of the rule for a composition. Let \(F\) be differentiable at \(u=g(a)\), and let \(g\) be differentiable at \(a\). Differentiability of \(g\) means that for a small change \(h\), its change in output can be written as
Write \(k=g(a+h)-g(a)\). Differentiability of \(F\) at \(u\) means
Substituting \(u=g(a)\) and \(k=g(a+h)-g(a)\) gives the change in the composition. Since \(k=g'(a)h+o(h)\), we have \(k=O(h)\), so a remainder that is \(o(k)\) is also \(o(h)\) as \(h\) tends to zero. (If \(k=0\), the change in \(F\) is zero as well.) Therefore
Divide by \(h\) and take the limit as \(h\) tends to zero. The term \(o(h)/h\) tends to zero, leaving \((F\circ g)'(a)=F'(g(a))g'(a)\). To obtain an inverse trigonometric composition formula, take \(F\) to be the required inverse trigonometric function and substitute its derivative formula for \(F'\). This explains both why the inner derivative is required and why the outer formula is evaluated at \(g(x)\).
Worked Examples
Worked Example: Differentiate an Arcsine Composition
Differentiate \(y=\arcsin\left(\frac{x+1}{3}\right)\), then find \(y'\) at \(x=0\).
Set \(g(x)=\frac{x+1}{3}\). Its derivative is \(g'(x)=\frac{1}{3}\). The input condition for the arcsine derivative is \(\left|\frac{x+1}{3}\right|<1\), which holds at \(x=0\), since the input there is \(\frac{1}{3}\). Apply the arcsine composition rule:
At \(x=0\), this gives \(y'=\frac{1}{\sqrt{9-1}}=\frac{1}{\sqrt{8}}=\frac{1}{2\sqrt{2}}\). Check using the unsimplified chain rule expression: \(\frac{1/3}{\sqrt{1-(1/3)^2}}=\frac{1/3}{\sqrt{8/9}}=\frac{1/3}{2\sqrt{2}/3}=\frac{1}{2\sqrt{2}}\). The two calculations agree.
Worked Example: Differentiate an Arctangent Composition
Differentiate \(y=\arctan(x^2-1)\), then evaluate the derivative at \(x=2\).
Let \(g(x)=x^2-1\), so \(g'(x)=2x\). The arctangent derivative formula applies for every real input. Thus
At \(x=2\), the inner value is \(g(2)=2^2-1=3\), and the inner derivative is \(g'(2)=2(2)=4\). Therefore \(y'(2)=\frac{4}{1+3^2}=\frac{4}{10}=\frac{2}{5}\). Check by substituting these values into the composition rule: the outer derivative at input \(3\) is \(1/(1+3^2)=1/10\), and multiplication by the inner derivative \(4\) gives \(4/10=2/5\).
Worked Example: Differentiate an Arcsecant Composition
Differentiate \(y=\arcsec(2x+3)\), then evaluate the derivative at \(x=0\).
Here \(g(x)=2x+3\), so \(g'(x)=2\). At \(x=0\), \(g(0)=3\), which satisfies the arcsecant condition \(|g(0)|>1\). In a neighborhood of \(0\), the input remains greater than \(1\), so the composition is differentiable there. Use the arcsecant formula, retaining the absolute value:
At \(x=0\), this becomes \(y'(0)=\frac{2}{|3|\sqrt{3^2-1}}=\frac{2}{3\sqrt{8}}=\frac{2}{6\sqrt{2}}=\frac{1}{3\sqrt{2}}\). Check the factors separately: the derivative of arcsecant at input \(3\) is \(1/(3\sqrt{8})\), and the inner derivative is \(2\). Their product is \(2/(3\sqrt{8})=1/(3\sqrt{2})\), as obtained above.
Worked Example: Keep Track of the Negative Sign for Arccosine
Differentiate \(y=\arccos\left(1-\frac{x}{2}\right)\), then find \(y'\) at \(x=1\).
Set \(g(x)=1-\frac{x}{2}\), giving \(g'(x)=-\frac{1}{2}\). At \(x=1\), \(g(1)=\frac{1}{2}\), which is strictly between \(-1\) and \(1\). The arccosine derivative includes a negative sign:
At \(x=1\), \(y'(1)=\frac{1}{2\sqrt{1-(1/2)^2}}=\frac{1}{2\sqrt{3/4}}=\frac{1}{\sqrt{3}}\). For a check, the outer derivative at input \(1/2\) is \(-1/\sqrt{1-1/4}=-2/\sqrt{3}\); multiplying by the inner derivative \(-1/2\) gives \(1/\sqrt{3}\). The two negative signs cancel.
A Reliable Differentiation Process
For a new composition, identify the inner function before differentiating. Then use the corresponding inverse trigonometric derivative formula and multiply by the derivative of the inner function. Keep the input condition in view, especially when using arcsine, arccosine, or arcsecant.
For \(F(g(x))\), write down \(g(x)\) and calculate \(g'(x)\) separately.
Use the formula for the inverse trigonometric function, evaluated at \(g(x)\).
Include all factors and signs from \(g'(x)\).
Confirm the input condition at the point or interval where the derivative is being used, then substitute values carefully.
A Common Mistake: Omitting the Inner Derivative
For \(y=\arcsin(3x)\), an incorrect calculation is
This uses the arcsine derivative formula but treats the input \(3x\) as if its derivative were \(1\). In fact, \(g(x)=3x\) has derivative \(g'(x)=3\), so the correct calculation is
A check at \(x=0\) makes the error clear. The incorrect expression gives \(1\), whereas the chain rule gives \(3/\sqrt{1}=3\). At this point the outer derivative is \(1\), and the inner derivative is \(3\); their product must be \(3\). The missing factor is not optional: it records how quickly the input to the arcsine is changing.
Why These Rules Matter
Inverse trigonometric functions often appear with inputs that are themselves expressions, not just \(x\). The chain rule makes their derivatives usable in those settings without requiring a new formula for every possible input. Once the outer derivative and the inner derivative are kept distinct, the calculation follows a consistent pattern.
Careful use of the conditions matters just as much as careful multiplication. An algebraically correct expression does not establish that the derivative exists at a point where the relevant inverse trigonometric function is not differentiable. Check the input condition, preserve signs and absolute values, and then evaluate.
Check Your Understanding
For each question, identify the inner function before applying the relevant inverse trigonometric derivative formula.
- Differentiate \(y=\arcsin(2x-1)\). What condition on \(x\) allows you to use the displayed arcsine derivative formula?
- Differentiate \(y=\arctan(4-x^2)\), and evaluate \(y'\) at \(x=1\).
- Differentiate \(y=\arccos(3x)\). Which two negative signs or factors must be accounted for?
- Differentiate \(y=\arcsec(x+4)\) at \(x=0\), retaining the absolute value in the formula.
- Explain why differentiating \(\arcsin(g(x))\) requires a factor of \(g'(x)\).
