Advanced Differentiation Proof Workshop

Turning an Inverse Difference Quotient Around

Differentiation and Approximation used the Mean Value Theorem to turn control of a derivative into control of function values. A related proof technique works in the opposite direction: when a function has an inverse, rewrite the inverse’s difference quotient as the reciprocal of a difference quotient for the original function. This simple identity is the core of the inverse derivative formula.

For an inverse \(g=f^{-1}\), let \(y_0=f(a)\), so \(g(y_0)=a\). If \(y\ne y_0\) and \(x=g(y)\), then \(x\ne a\), and \(f(x)=y\). Consequently,

The expression on the right resembles the reciprocal of \(f'(a)\). But the resemblance alone is not a proof: as \(y\) approaches \(y_0\), we must know that \(x=g(y)\) approaches \(a\). That is a continuity question about the inverse. Strict monotonicity and continuity of \(f\) provide the needed link.

The Inverse Derivative Formula

Proof. First we verify continuity of \(g\) at \(y_0\). Fix \(\varepsilon>0\). Since \(I\) is open, choose \(r\) with \(0<r<\varepsilon\) such that \(a-r,a+r\in I\). Strict monotonicity implies that \(f(a-r)\) and \(f(a+r)\) lie on opposite sides of \(f(a)\). Thus

If \(y\in f(I)\) and \(|y-y_0|<\eta\), monotonicity forces \(g(y)\) to lie between \(a-r\) and \(a+r\). Hence \(|g(y)-a|<r<\varepsilon\), proving that \(g(y)\to a\) as \(y\to y_0\) within \(f(I)\).

Now take \(y\ne y_0\) in \(f(I)\), and put \(x=g(y)\). Because \(f\) is one-to-one, \(x\ne a\). The inverse identity gives

As \(y\to y_0\), the continuity just proved gives \(x=g(y)\to a\). Since \(f\) is differentiable at \(a\), the denominator in the final expression tends to \(f'(a)\), which is nonzero. Taking the limit therefore yields

This proves the formula. The nonzero derivative hypothesis ensures that the reciprocal has a finite limit. \(\square\)

The proof has two distinct jobs. Strict monotonicity gives a well-defined inverse and ensures that inputs approaching \(f(a)\) correspond to original inputs approaching \(a\). Differentiability and the nonzero derivative then determine the limiting reciprocal slope. Keeping these jobs separate is a useful proof habit: an algebraic rearrangement of quotients cannot replace the continuity argument.

A Continuously Differentiable Inverse

The pointwise theorem gives a derivative at one output value. With continuous derivative information throughout an interval, the same argument applies at every point and also shows that the inverse derivative varies continuously.

Proof. The derivative \(f'\) is continuous and never zero. It cannot take both positive and negative values on \(I\), because the Intermediate Value Theorem would then give a point where it is zero. Thus \(f'\) has a constant sign. For any \(u<v\) in \(I\), the Mean Value Theorem gives a point \(c\in(u,v)\) such that

Since \(v-u>0\) and \(f'(c)\) has the same nonzero sign throughout \(I\), \(f(v)-f(u)\) has that sign. Hence \(f\) is strictly monotone and has an inverse on \(f(I)\).

Fix any \(y\in f(I)\), and let \(a=g(y)\). The inverse derivative theorem applies at \(a\), giving \(g'(y)=1/f'(a)=1/f'(g(y))\). This proves the stated formula at every \(y\in f(I)\). The inverse \(g\) is continuous by the continuity argument in the preceding theorem. Since \(f'\) and \(g\) are continuous and \(f'\) never vanishes, the function \(y\mapsto1/f'(g(y))\) is continuous. The derivative \(g'\) is therefore continuous on \(f(I)\). \(\square\)

This result is useful when an inverse must be differentiated repeatedly or when continuity of its slope matters. It also makes the domain of the formula explicit: the derivative of the inverse at \(y\) uses the derivative of the original function at the corresponding point \(g(y)\), not at \(y\) itself.

Why the Nonzero Derivative Condition Matters

The reciprocal formula cannot be used when \(f'(a)=0\). This is not merely a technical obstacle: the inverse may fail to have a finite derivative there. For instance, the inverse of \(x^3\) is a cube root, whose difference quotient at zero becomes unbounded.

There is a subtler pitfall: \(f'(a)\ne0\) at a single point does not by itself guarantee that \(f\) has an inverse on a whole neighborhood of that point. For example, define \(F(0)=0\) and \(F(x)=x+x^2\sin(1/x^2)\) for \(x\ne0\). The derivative at zero is

But for \(x\ne0\),

Set \(x_n=1/\sqrt{2\pi n}\). Then \(\sin(1/x_n^2)=0\), \(\cos(1/x_n^2)=1\), and \(F'(x_n)=1-2/x_n\), which is negative for all sufficiently large \(n\). Such points occur arbitrarily close to zero. If \(F\) were one-to-one on an interval around zero, its continuity would force it to be strictly monotone there. It cannot be decreasing, since \(F'(0)=1\); and it cannot be increasing, since an increasing differentiable function has nonnegative derivative at every interior point. Thus \(F\) is not one-to-one on any interval around zero.

The inverse derivative theorem avoids this problem by assuming strict monotonicity on an interval, rather than inferring local invertibility from one nonzero derivative value. A practical proof should check the inverse exists first, establish continuity of that inverse at the point of interest, and only then take reciprocals of difference quotients.