One Relationship, Several Kinds of Information
Inverse-function derivative questions can look quite different depending on what information is provided. An equation may give the function directly, a table may list its values and derivatives, or a graph may show the relevant tangent. A composition may involve an inverse trigonometric function and the chain rule.
In every case, the central task is to keep the inputs and outputs straight. If \(f(a)=b\), then the inverse takes \(b\) back to \(a\). When the inverse derivative exists and \(f'(a)\ne 0\), its value at \(b\) is the reciprocal of the original derivative at \(a\).
The formula requires the inverse to exist on the relevant interval and the original derivative at the corresponding input to be nonzero. A zero derivative cannot be put into the denominator; it signals that this formula does not give a finite inverse derivative there.
Why the Reciprocal-Slope Rule Works
The inverse-derivative formula can be justified without assuming in advance that the inverse is differentiable. Let \(f\) be differentiable near \(a\), one-to-one on an interval around \(a\), and suppose \(f'(a)\ne 0\). Write \(b=f(a)\) and let \(g=f^{-1}\). Because a continuous one-to-one function on an interval has a continuous inverse on its image, \(g(y)\to a\) as \(y\to b\).
For \(y\ne b\), set \(x=g(y)\). Then \(y=f(x)\), and \(x\to a\) as \(y\to b\). The difference quotient for \(g\) can be rewritten using the inverse relationship:
As \(y\to b\), \(x\to a\), so the denominator in the final expression tends to \(f'(a)\). Since \(f'(a)\ne 0\), taking reciprocals is valid, and therefore
This proof also explains why the input correspondence matters: the limiting derivative of \(f\) is taken at \(a=g(b)\), not at \(b\) unless those happen to be the same number.
Worked Example: Finding the Corresponding Input from an Equation
Worked Example: Finding the Corresponding Input from an Equation
Let \(f(x)=x^3+x+1\). Find \((f^{-1})'(3)\), and verify the result.
First identify the input \(a\) for which \(f(a)=3\). Testing \(a=1\) gives
Thus \(f^{-1}(3)=1\), so the original derivative must be evaluated at \(x=1\). Differentiate \(f\):
Because \(f'(1)=4\ne 0\), the inverse-derivative formula applies at this point. Hence
Check: The corresponding original slope is \(4\), and the claimed inverse slope is \(\frac14\). Their product is \(4\cdot\frac14=1\), as required by the reciprocal-slope relationship.
It is useful to write the correspondence explicitly before differentiating: \(f(1)=3\), so the inverse input \(3\) corresponds to the original input \(1\). This small step prevents a frequent substitution error.
Worked Example: Reading a Table
Worked Example: Reading a Table
A table gives \(f(2)=5\) and \(f'(2)=-3\). Assuming \(f\) is one-to-one near \(x=2\), find \((f^{-1})'(5)\) and check it.
The table says that the original input \(2\) produces the output \(5\). Therefore \(f^{-1}(5)=2\), and the original derivative needed is \(f'(2)\), which the table gives as \(-3\). Since this is nonzero,
Check: The original slope \(-3\) multiplied by the inverse slope \(-\frac13\) gives \((-3)(-\frac13)=1\). The negative sign is necessary: a decreasing function has a decreasing inverse on the corresponding interval.
A table may list several inputs and outputs. Read the row or column containing the inverse’s input as an output of the original function. Then use the derivative listed at the original input that produced it.
Worked Example: Reading a Tangent from a Graph
Worked Example: Reading a Tangent from a Graph
Suppose a graph of \(f\) passes through \((2,-1)\), and the tangent line there passes through \((0,-4)\). Find \((f^{-1})'(-1)\), assuming \(f\) is one-to-one near \(x=2\), and verify the result.
The point \((2,-1)\) means \(f(2)=-1\). Thus the inverse input \(-1\) corresponds to original input \(2\). The tangent slope at \((2,-1)\) is found from the two points on the tangent line:
This slope is nonzero, so the inverse derivative is
Check: The graph of the inverse reflects the graph of \(f\) across \(y=x\). The corresponding tangent slope is therefore the reciprocal, and \(\frac32\cdot\frac23=1\).
When extracting a slope from a graph, use two points on the tangent line, not just the point where the curve is evaluated. The point on the curve gives the input-output correspondence; the tangent gives the derivative.
Worked Example: Combining Inverse-Trigonometric Compositions
Worked Example: Combining Inverse-Trigonometric Compositions
Let \(H(x)=\arcsin(x)+\arctan(2x)\). Find \(H'(\frac12)\), and verify both terms.
For \(-1<x<1\), \(\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}\). For any real \(u\), \(\frac{d}{dx}\arctan(u)=\frac{u'}{1+u^2}\). Apply these formulas and the chain rule to the second term:
At \(x=\frac12\), the arcsine term is
The arctangent term is
Therefore
Check: For the arcsine term, the inner input is \(u=x\), with \(u'=1\); its derivative is \(u'/\sqrt{1-u^2}=2/\sqrt3\) at \(x=\frac12\). For the arctangent term, the inner input is \(u=2x\), with \(u'=2\); its derivative is \(u'/(1+u^2)=2/2=1\). These reproduce both terms in the result. The arcsine input \(\frac12\) lies strictly between \(-1\) and \(1\), so its derivative formula applies.
Inverse-trigonometric derivatives are often compositions, so the chain-rule factor is essential. For example, differentiating \(\arctan(2x)\) as though its input were simply \(x\) would omit the factor \(2\).
A Reliable Process for Mixed Problems
The same short routine works across equations, tables, graphs, and inverse-trigonometric formulas. It separates the correspondence question from the derivative calculation.
If the problem asks for \((f^{-1})'(b)\), record \(b\).
Locate \(a\) such that \(f(a)=b\), using the equation, table, or graph.
Determine \(f'(a)\), taking care to use the derivative at \(a\), not at \(b\).
Confirm \(f'(a)\ne0\), then calculate \((f^{-1})'(b)=1/f'(a)\).
Multiply the original and inverse slopes; the product should equal \(1\).
Common Mistake: Reversing the Evaluation Point
Suppose \(f(2)=5\) and \(f'(2)=4\), and the question asks for \((f^{-1})'(5)\). A common incorrect calculation is
This uses the inverse input \(5\) as though it were the original input. The equation \(f(2)=5\) shows that the original input corresponding to \(5\) is \(2\), so the corrected calculation is
The check confirms the correction: the original slope \(4\) times the inverse slope \(\frac14\) equals \(1\). Always write \(f(a)=b\) before using the formula; it makes the two evaluation points visible.
When the Formula Does Not Apply
Do not use the reciprocal formula without checking its assumptions. If \(f\) is not one-to-one on the relevant domain, there may be no single-valued inverse there; a restricted branch may be needed. If \(f'(a)=0\), division by \(f'(a)\) is not valid, and the inverse derivative may fail to exist as a finite number. Endpoint cases for inverse trigonometric functions also require attention to the domain and endpoint behavior; do not substitute into an interior derivative formula without checking that it applies.
For inverse-trigonometric compositions, also check the input to the inverse function and retain the chain-rule factor. For instance, \(\arcsin(u(x))\) requires \(-1<u(x)<1\) for the usual finite derivative formula at an interior point, and
These checks are not extra formalities. They tell you whether the expression being calculated is defined and whether the derivative rule is applicable.
Check Your Understanding
For each question, identify the input-output correspondence before calculating.
- If \(f(3)=7\) and \(f'(3)=-2\), find \((f^{-1})'(7)\), assuming the inverse exists near the corresponding point.
- A graph shows \(f(-1)=4\), with tangent slope \(f'(-1)=\frac52\). Find the derivative of the inverse at \(4\).
- If \(f(x)=x^2+x\) on a domain where it is one-to-one, and \(f(2)=6\), calculate \((f^{-1})'(6)\) and verify by multiplying slopes.
- Differentiate \(K(x)=\arctan(3x)\) and state the chain-rule factor.
- Why is it incorrect to use \(1/f'(b)\) automatically when asked to find \((f^{-1})'(b)\)?
