已知函数f及其导数f'，导数的反函数(f')⁻¹是否有通用推导公式？

Great questions—let’s unpack this step by step, since these are common points of confusion when working with inverse functions and calculus.

1. Is there a universal formula for the inverse of a derivative? #

Short answer: No, there’s no general closed-form formula that works for all derivatives. The ability to write the inverse of ( f' ) explicitly depends entirely on the form of ( f' ) itself.

• For simple derivatives (linear, exponential, basic trigonometric/inverse trigonometric functions), you can often derive an explicit inverse using standard algebraic manipulation. For example:

  • If ( f(x) = e^x + x ), then ( f'(x) = e^x + 1 ). Solving ( y = e^x + 1 ) for ( x ) gives ( x = \ln(y - 1) ), which is the explicit inverse ( (f')^{-1}(y) = \ln(y - 1) ).
  • If ( f(x) = \tan(x) ) (defined on ( (-\pi/2, \pi/2) )), then ( f'(x) = \sec^2(x) = 1 + \tan^2(x) ). The inverse here is ( (f')^{-1}(y) = \arctan(\sqrt{y - 1}) ) for ( y > 1 ).

• For most non-trivial derivatives, however, you can’t express the inverse using elementary functions. Take ( f(x) = x^5 + x ): its derivative is ( f'(x) = 5x^4 + 1 ), which is strictly increasing (so invertible), but there’s no way to write ( x ) in terms of ( y ) (where ( y = 5x^4 + 1 )) using basic arithmetic, exponents, logs, or trig functions. In these cases, you’d rely on implicit notation (( 5[(f')^{-1}(y)]4 + 1 = y )) or numerical methods (like Newton-Raphson) to approximate values of ( (f')^{-1}(y) ).

2. If ( (f')^{-1} ) exists, is there a universal method to derive it? #

First, recall that ( (f')^{-1} ) exists if and only if ( f' ) is strictly monotonic (for continuous derivatives, which is most cases we work with, strict monotonicity is equivalent to invertibility). This usually means ( f''(x) ) doesn’t change sign (so ( f ) is either strictly convex or strictly concave over its domain).

While there’s no one-size-fits-all closed-form formula, there is a general process to find ( (f')^{-1} ):

• Start with the defining equation of the inverse: ( f'\left( (f')^{-1}(y) \right) = y ).
• Treat this as an equation where you need to solve for ( (f')^{-1}(y) ) in terms of ( y ).

The success of this process depends on how solvable the equation ( f'(x) = y ) is. For example:

• If ( f'(x) = 3x^2 ) (for ( x > 0 ), where it’s strictly increasing), solving ( y = 3x^2 ) gives ( x = \sqrt{y/3} ), so ( (f')^{-1}(y) = \sqrt{y/3} ).
• If ( f'(x) = \sin(x) ) (restricted to ( [-\pi/2, \pi/2] ) where it’s strictly increasing), the inverse is just ( (f')^{-1}(y) = \arcsin(y) ), a standard inverse trigonometric function.

Additionally, if you need the derivative of ( (f')^{-1} ) (not the inverse itself), there’s a standard formula:

[(f')^{-1}]'(y) = 1 / f''\left( (f')^{-1}(y) \right)

This holds as long as ( f''\left( (f')^{-1}(y) \right) \neq 0 ), and it’s derived directly from the chain rule for inverse functions.

内容的提问来源于stack exchange，提问作者Thomas Eberhard