Why the Second Derivative of an Inverse Matters
An inverse function reverses the roles of input and output. If \(f(a)=x\), then \(f^{-1}(x)=a\). This reversal affects derivatives: the rate of change of the inverse depends on the rate of change of the original function. For the second derivative, both the first and second derivatives of the original function matter.
The key is to keep track of which input is being used. To find the second derivative of \(f^{-1}\) at \(x\), first find the number \(a\) for which \(f(a)=x\). Then evaluate \(f'(a)\) and \(f''(a)\). The formula is
The cube in the denominator and the negative sign are both essential. We will derive them from the identity that defines an inverse, then use the formula in several settings.
Conditions and Notation
Suppose \(f\) has a local inverse near \(a\), and let \(g=f^{-1}\) denote that inverse. A sufficient standard condition is that \(f\) is twice continuously differentiable near \(a\) and \(f'(a)\ne 0\). Under this condition, the inverse function theorem guarantees a local inverse with the derivatives needed below. A function that is one-to-one on a larger interval may also have an inverse there; the formula applies wherever the stated derivatives exist and \(f'\) is nonzero.
The notation \(f''(a)\) means the second derivative of the original function evaluated at its input \(a\). It does not mean the second derivative of the inverse evaluated at \(a\). The inverse’s input is \(x=f(a)\), while the original function’s input is \(a\). Matching those two points correctly is a central part of using the formula.
Deriving the Formula from the Inverse Identity
The identity \(f(g(x))=x\) expresses that applying \(g\) and then \(f\) returns the starting value \(x\). Differentiating this identity once gives the first-derivative formula. Differentiating it a second time produces the formula for the inverse’s second derivative.
For \(g=f^{-1}\), write \(f(g(x))=x\).
The chain rule gives \(f'(g(x))g'(x)=1\), so \(g'(x)=1/f'(g(x))\).
Apply the product and chain rules to \(f'(g(x))g'(x)=1\), then solve for \(g''(x)\).
Here is the full second differentiation. On the left side of \(f'(g(x))g'(x)=1\), the product rule gives
The derivative of the right side, \(1\), is \(0\). Therefore,
We already know from the first differentiation that \(g'(x)=1/f'(g(x))\). Substitute this expression:
Since \(f'(g(x))\ne 0\), we can isolate \(g''(x)\). First move the first term to the right, then divide by \(f'(g(x))\):
Finally, at the point \(x=f(a)\), the inverse identity gives \(g(x)=a\). Replacing \(g(x)\) by \(a\) yields \(g''(x)=-f''(a)/[f'(a)]^3\). This derivation explains the cube: two factors of \(f'\) arise when substituting for \([g'(x)]^2\), and one more arises when dividing by \(f'(g(x))\).
Worked Example: A Polynomial with an Increasing Inverse
Worked Example: Find an Inverse Second Derivative from Polynomial Data
Let \(f(t)=t^3+t\). Find \((f^{-1})''(2)\), and verify the result using the inverse derivative relation. First, find the corresponding original input: \(f(1)=1^3+1=2\), so \(a=1\) and \(f^{-1}(2)=1\).
Differentiate the original function:
Evaluate at the corresponding input \(a=1\):
Now apply the formula:
To check, write \(g=f^{-1}\). The first-derivative formula, valid at every input \(x\), is \(g'(x)=1/(3[g(x)]^2+1)\). Differentiating it by the chain rule gives
At \(x=2\), \(g(2)=1\), so this gives \(g''(2)=-6/(3(1)^2+1)^3=-6/64=-3/32\), matching the formula.
Worked Example: An Exponential Function
Worked Example: Find and Check the Second Derivative of a Logarithmic Inverse
Let \(f(t)=e^t\), whose inverse is \(g(x)=\ln x\) for \(x>0\). Find \(g''(1)\) using the inverse-function formula. The corresponding original input is \(a=0\), because \(f(0)=e^0=1\).
The first two derivatives of \(f\) are \(f'(t)=e^t\) and \(f''(t)=e^t\). At \(a=0\),
Thus
Check by differentiating the inverse directly: \(g(x)=\ln x\), so \(g'(x)=1/x\) and \(g''(x)=-1/x^2\). At \(x=1\), \(g''(1)=-1/1^2=-1\), as required.
Worked Example: A Decreasing Inverse Branch
Worked Example: Use the Formula When the Original Function Is Decreasing
Let \(f(t)=t^2-2t+5\), restricted to the interval \((-1/2,1/2)\). Find the second derivative of this inverse branch at \(x=5\). The restriction matters: the quadratic is not one-to-one on all real numbers, but it is strictly decreasing on this interval, since \(f'(t)=2t-2<0\) there. The corresponding input is \(a=0\), because \(f(0)=5\).
The needed derivative values are
Apply the formula, keeping the negative value of \(f'(0)\) inside the cube:
Check by differentiating the inverse relation in implicit form. Write \(y=g(x)\), so \(y^2-2y+5=x\). Differentiating with respect to \(x\) gives \((2y-2)y'=1\), hence \(y'=1/(2y-2)\). Differentiate the relation \((2y-2)y'=1\) again:
At \(x=5\), \(y=0\), and the first derivative equation gives \(y'=1/(2(0)-2)=-1/2\). Substituting both values into the second derivative equation gives \(2(-1/2)^2+(-2)y''=0\), so \(1/2-2y''=0\) and \(y''=1/4\). This agrees with the formula.
How to Apply the Formula Reliably
The calculation is short once the correct corresponding input is identified. A reliable solution keeps the input and output roles visible rather than substituting a number into the wrong function.
For \((f^{-1})''(x)\), determine \(a\) such that \(f(a)=x\).
Confirm that the inverse branch is defined and that \(f'(a)\ne 0\).
Find \(f'(a)\) and \(f''(a)\), not values at the inverse input unless those happen to be the same.
Compute \(-f''(a)/[f'(a)]^3\), preserving the sign of \(f'(a)\).
The condition \(f'(a)\ne 0\) is not a minor technical detail. The first derivative of the inverse is \(1/f'(a)\); if \(f'(a)=0\), that expression is undefined, so this theorem does not provide an inverse derivative there. A particular inverse may still have some special behavior, but it must be analyzed separately rather than by using this formula.
Common Pitfall: Forgetting the Cube
A frequent error is to use a square in the denominator, perhaps because the second differentiation produces \([g'(x)]^2\). That square is only one part of the calculation. After substituting for \(g'(x)\), we must also divide by \(f'(g(x))\) to isolate \(g''(x)\). The resulting denominator is cubed.
For the polynomial example, using the incorrect square would give
which is not the correct value \(-3/32\). The correction is to use the derived formula with the cube:
Another related error is to discard the sign of \(f'(a)\). In the decreasing-branch example, \([f'(0)]^3=(-2)^3=-8\), not \(8\). Keeping the signed value gives a positive inverse second derivative, \(1/4\).
What the Formula Tells Us
The formula shows that the curvature of the inverse is influenced by two features of the original function at the corresponding input. The value \(f''(a)\) describes how the original function’s slope is changing, while \(f'(a)\) determines how strongly input and output scales are reversed. The sign and magnitude of the inverse’s second derivative therefore cannot be inferred from \(f''(a)\) alone.
For example, when \(f''(a)>0\), the numerator in the formula is negative, but the denominator can be positive or negative depending on the sign of \(f'(a)\). If \(f'(a)>0\), then \((f^{-1})''(x)<0\). If \(f'(a)<0\), then \((f^{-1})''(x)>0\). This conclusion follows because cubing preserves the sign of a nonzero number.
The formula is useful whenever a problem gives local derivative information about a function but asks about its inverse. It avoids finding an explicit expression for the inverse, which may be difficult or impossible to write in a convenient form. The essential data are the matching input \(a\), the slope \(f'(a)\), and the second derivative \(f''(a)\).
Check Your Understanding
For each question, identify the input of the original function that corresponds to the inverse input before calculating.
- If \(f(2)=6\), \(f'(2)=3\), and \(f''(2)=9\), find \((f^{-1})''(6)\).
- Suppose \(f(-1)=4\), \(f'(-1)=-2\), and \(f''(-1)=6\). Find \((f^{-1})''(4)\).
- Starting from \(f(g(x))=x\), explain why differentiating twice produces a term \(f'(g(x))g''(x)\).
- What condition on \(f'(a)\) is needed for the formula, and why does the derivation need it?
- A student uses \(-f''(a)/[f'(a)]^2\). Which step in the derivation has the student omitted?
