Introduction
An inverse function reverses the input-output relationship of a function. If \(f(a)=b\), then \(f^{-1}(b)=a\), provided the inverse is defined on the relevant range. Differentiating this reversal once gives a reciprocal-slope relationship. Differentiating again and a third time reveals how the curvature and higher-order behavior of \(f\) contribute to the derivatives of its inverse.
The key is to keep track of which variable is being differentiated. Write \(g=f^{-1}\), and set \(y=g(x)\). The defining identity is then \(f(y)=x\). Derivatives of \(f\) in the formulas below are evaluated at \(y=g(x)\), not at \(x\), unless the two happen to be equal.
Conditions and notation
The nonzero derivative condition matters: it ensures that \(f\) is locally invertible and that the reciprocal in the first-derivative formula is defined. For the second-derivative formula, \(f''\) must exist; for the third-derivative formula, \(f'''\) must exist. We use \(y=g(x)\), so \(f(y)=x\).
These expressions are useful when the inverse is difficult to write explicitly. To find its derivatives at a particular \(x\), first find \(y\) satisfying \(f(y)=x\), then evaluate the derivatives of \(f\) at that \(y\).
Deriving the inverse derivative formulas
Start with \(f(g(x))=x\). The chain rule gives the first derivative. To obtain the next formulas, differentiate the resulting identities with respect to \(x\), remembering that \(y=g(x)\) depends on \(x\).
From \(f(y)=x\), the chain rule gives \(f'(y)g'(x)=1\), so \(g'(x)=1/f'(y)\). This requires \(f'(y)\ne 0\).
Differentiate \(f'(y)g'(x)=1\) by the product and chain rules, then solve for \(g''(x)\).
Differentiate the second-derivative expression, using \(dy/dx=g'(x)=1/f'(y)\), and simplify.
Here is the full calculation for the second derivative. In the product rule, differentiating \(f'(y)\) with respect to \(x\) gives \(f''(y)g'(x)\):
Substitute \(g'(x)=1/f'(y)\), then isolate \(g''(x)\):
For the third derivative, differentiate \(g''(x)=-f''(y)(f'(y))^{-3}\). Applying the product and chain rules gives
The second line uses \(g'(x)=1/f'(y)\) in each term. This derivation also explains why the denominator in the third-derivative formula has the fifth power: differentiating with respect to \(x\) introduces another factor \(1/f'(y)\), in addition to the factors already present.
Worked examples
Worked Example: A cubic inverse at a specified point
Let \(f(t)=t+t^3\), and let \(g=f^{-1}\) near \(t=1\). Find \(g'(2)\), \(g''(2)\), and \(g'''(2)\).
First locate the corresponding input to \(f\). Since \(f(1)=1+1=2\), we have \(g(2)=1\). Also, $$ f'(t)=1+3t^2,\qquad f''(t)=6t,\qquad f'''(t)=6. $$ At \(t=1\), these values are \(f'(1)=4\), \(f''(1)=6\), and \(f'''(1)=6\). In particular \(f'(1)\ne0\), as required.
Apply the formulas, evaluating each derivative of \(f\) at \(g(2)=1\): $$ g'(2)=\frac{1}{4},\qquad g''(2)=-\frac{6}{4^3}=-\frac{6}{64}=-\frac{3}{32}, $$ $$ g'''(2)=\frac{3(6)^2-(4)(6)}{4^5} =\frac{108-24}{1024} =\frac{84}{1024} =\frac{21}{256}. $$
Check the values by differentiating the composition identity. At \(x=2\), the first-derivative identity gives \(f'(1)g'(2)=4(1/4)=1\). The second-derivative identity gives $$ f''(1)(g'(2))^2+f'(1)g''(2) =6\left(\frac14\right)^2+4\left(-\frac{3}{32}\right) =\frac{3}{8}-\frac{3}{8}=0. $$ For the third derivative, differentiating the second-derivative identity yields \(f'''(y)(g')^3+3f''(y)g'g''+f'(y)g'''=0\). At the point in question, $$ 6\left(\frac14\right)^3 +3(6)\left(\frac14\right)\left(-\frac{3}{32}\right) +4\left(\frac{21}{256}\right) =\frac{3}{32}-\frac{27}{64}+\frac{21}{64}=0. $$ All three values satisfy the differentiated composition identities.
Worked Example: An inverse with zero second derivative
Let \(f(t)=t+\sin t\), and consider its local inverse \(g\) near \(x=0\). Find the first three derivatives of \(g\) at \(0\).
We have \(f(0)=0\), so \(g(0)=0\). The derivatives are $$ f'(t)=1+\cos t,\qquad f''(t)=-\sin t,\qquad f'''(t)=-\cos t. $$ Thus \(f'(0)=2\), \(f''(0)=0\), and \(f'''(0)=-1\). Because \(f'(0)=2\ne0\), the inverse exists locally and the formulas apply.
Substitution gives $$ g'(0)=\frac12,\qquad g''(0)=-\frac{0}{2^3}=0, $$ $$ g'''(0)=\frac{3(0)^2-(2)(-1)}{2^5} =\frac{2}{32} =\frac{1}{16}. $$
To check, the first differentiated identity gives \(2(1/2)=1\). The second gives \(0(1/2)^2+2(0)=0\). The third gives $$ (-1)\left(\frac12\right)^3 +3(0)\left(\frac12\right)(0) +2\left(\frac{1}{16}\right) =-\frac18+\frac18=0. $$ Each composition identity is satisfied.
Worked Example: The inverse of the natural logarithm
Let \(f(t)=\ln t\) for \(t>0\). Its inverse is \(g(x)=f^{-1}(x)\). Find \(g'(0)\), \(g''(0)\), and \(g'''(0)\) using the inverse formulas.
Because \(f(1)=\ln 1=0\), the corresponding input is \(g(0)=1\). The derivatives of \(f\) are $$ f'(t)=\frac1t,\qquad f''(t)=-\frac1{t^2},\qquad f'''(t)=\frac2{t^3}. $$ At \(t=1\), these are \(1\), \(-1\), and \(2\), respectively, and \(f'(1)\ne0\).
Therefore $$ g'(0)=\frac11=1,\qquad g''(0)=-\frac{-1}{1^3}=1, $$ $$ g'''(0)=\frac{3(-1)^2-(1)(2)}{1^5} =\frac{3-2}{1}=1. $$
For a check, the inverse here is \(g(x)=e^x\), whose first, second, and third derivatives at \(0\) are all \(e^0=1\). We can also check the composition identities: the first gives \(1\cdot1=1\), the second gives \((-1)(1)^2+(1)(1)=0\), and the third gives \(2(1)^3+3(-1)(1)(1)+(1)(1)=2-3+1=0\). The inverse formulas agree with both checks.
Common mistakes and how to avoid them
A frequent error is to evaluate derivatives of \(f\) at the output \(x\), rather than at the corresponding input \(y=g(x)\). For instance, in the cubic example, the requested inverse derivatives were at \(x=2\), but the derivatives of \(f\) had to be evaluated at \(y=1\), because \(f(1)=2\).
Another common error is losing a power or a minus sign. For the cubic example, a wrong second-derivative calculation might use $$ g''(2)=-\frac{f''(1)}{(f'(1))^2} =-\frac{6}{16} =-\frac38. $$
The denominator must be \((f'(y))^3\), not its square. The correct result is $$ g''(2)=-\frac{6}{4^3}=-\frac{3}{32}. $$ A useful check is the twice-differentiated identity: with the incorrect value, \(6(1/4)^2+4(-3/8)=3/8-3/2\ne0\), so the composition identity fails.
For the third derivative, preserve the entire numerator \(3(f'')^2-f'f'''\). It is not enough to differentiate only the numerator informally or to omit the factor introduced by \(dy/dx\). Keeping \(y=g(x)\) visible throughout the calculation helps prevent both errors.
Why higher inverse derivatives matter
The first derivative describes the slope of the inverse; it is the reciprocal of the original slope at the corresponding point. The second derivative describes how that inverse slope changes, and the third derivative records how the second derivative changes. The formulas show that inverse curvature depends not only on \(f''\), but also on the size and sign of \(f'\). The third derivative combines \(f''\) and \(f'''\), with a fifth power of the original slope in the denominator.
These relationships are especially useful when \(f^{-1}\) has no convenient explicit formula. One can still compute derivatives at a point by solving \(f(y)=x\) and evaluating \(f'(y)\), \(f''(y)\), and \(f'''(y)\). The local condition \(f'(y)\ne0\) remains essential; if it fails, these formulas cannot be used at that point.
Check Your Understanding
For each question, keep the inverse input \(y=g(x)\) distinct from the point \(x\) where the inverse derivative is requested.
- State the condition on \(f'\) needed to use the inverse derivative formulas locally.
- Write the formula for \(g''(x)\) in terms of derivatives of \(f\) at \(y=g(x)\).
- If \(f(2)=5\), which input to \(f\) is used to evaluate the derivatives when finding \(g'(5)\)?
- For \(f(t)=t^3+t\), use the formula to find \(g'(2)\), where \(g=f^{-1}\) near \(1\).
- Explain why differentiating with respect to \(x\) introduces a factor of \(g'(x)\) when differentiating an expression involving \(y=g(x)\).
