The relations between the properties of monotonicity, continuity and derivation of a function of one real variable allow us to formally derivate the inverse bijection of an injective and derivable function. The most representative example is perhaps that of the exponential function, the reciprocal of the neperian logarithm.

1 Monotonicity, continuity and derivation

In the elementary theory of real functions, one studies the subtle relationships between monotone, continuous and derivable functions, in close connection with the “topology” of the real line, i.e. the intervals where these functions are defined. Recall that an interval of the set $\mathbb R$ is a subset $I\subseteq \mathbb R$ such that for all $a, b\in I$, the segment $[a,b]=\{x\in \mathbb R : a\leq x\leq b\}$ is entirely included in $I$.

For example, it is proved that if $f:I\to\mathbb R$ is a continuous function defined on an interval $I$, then the image $f(I)$ of the function $f$ is also an interval. We also know that such an application is injective if and only if it is strictly monotonic (i.e. strictly increasing or strictly decreasing)! And in this case, when the interval $I$ is open, the image $f(I)$ is also an open interval. Now, whenever possible, we study the variations of such a function using its derivative (see What is the derivative of a function? Definition and geometric interpretation): a derivable function $f:I\to\mathbb R$ defined on an open interval $I$ is monotonic on $I$ if and only if $f'(t)$ keeps a constant sign on $I$, positive if $f$ is increasing, negative if $f$ is decreasing.

Using the definition of the derivative of a function, we can also derivate a compound function (see Wikipedia, for example: Derivation of compound functions). If $f:I\to \mathbb R$ and $g:J\to \mathbb R$ are two functions defined on open intervals $I$ and $J$, and derivable, and if $f(I)\subseteq J$, then the function $g\circ f$ is defined and we have $(g\circ f)'(x)=g'(f(x)). f'(x)$ for all $ \in I$.

2 Derivating the inverse bijection

From these properties we can easily obtain the expression for the derivative of an inverse bijection in certain cases. If $f:I\to \mathbb R$ is a differentiable and injective function defined on an open interval $I$, $f$ is continuous, so its image $f(I)$ is an open interval $J$. Considering it as an application from $I$ into $J$, let us denote $g:J\to I$ as its reciprocal bijection. By definition, we have $f\circ g(x)=x$ for all $x\in J$, which is the “identical” function $x\in J\mapsto x\in J$. If $g$ is differentiable, the derivative of $f\circ g$ is therefore the derivative of the identical function, which is $1$, and by the derivation formula for compound functions we can therefore write \[1=(f\circ g)'(x)=f'(g(x)).g'(x)\] for all $x\in J$.

In fact, in this situation and with the same notations we can show that the function $g:J\to I$ is then differentiable at any point $x\in J$ for which $f'(g(x))\neq 0$ and at this point, the previous formula – i.e. $1=(f\circ g)'(x)=f'(g(x)).g'(x)$ – gives us the derivative $$g'(x)=\dfrac{1}{f'(g(x))}.$$ In particular, if $f:I\to\mathbb R$ is derivable and strictly monotonic, and if its derivative $f’$ never cancels, then its reciprocal bijection $g:J\to I$ is derivable at any point $x\in J$ and the previous formula gives us a general expression for its derivative at any point.

3.Example: Derivating the exponential function

As an application of the derivation formula for an inverse bijection, we will show that the derivative of the exponential function is equal to the function itself, i.e. for all $x\in\mathbb R$, we have $$\exp'(x)=\exp(x).$$.
To do this, let us recall the usual definition of the exponential function. Using the properties of the Riemann integral, we first define the (neperian) logarithm function $\ln:\mathbb R_+^*\to \mathbb R$, by the expression $$\ln(x)=\int_1^x \dfrac 1 t \ dt$$ for any real number $x>0$. In other words, the Logarithm is the primitive of the function $x\in \mathbb R_+^*\mapsto 1/x$ which cancels at $x=1$. We then show that this function is strictly increasing and that $\limits_{x\to 0} \ln(x)=-\infty$ and $\limits_{x\to +\infty} \ln(x)=+\infty$, so as $\ln$ is by definition differentiable, with derivative $1/x$, it is a bijection from $\mathbb R_+^*$ onto $\mathbb R$. By definition, the exponential function is then the reciprocal bijection of the Logarithm, that is the unique mapping $\exp:\mathbb R\to \mathbb R_+^*$$ such that for all $t\in \mathbb R_+^*$ and $x\in \mathbb R$, we have $\exp(\ln(t))=t$ and $\ln(\exp(x))=x$. Let us then apply the stated theorem and the formula written previously with $f=\ln$, $I=\mathbb R_+^*$, $g=\exp$ and $J=\mathbb R$. Since $\ln$ is strictly increasing and differentiable, and since $\ln'(t)=1/t>0$ for all $t>0$, its reciprocal $\exp$ is differentiable, and its derivative at any point $xin \mathbb R$ is given by $$\exp'(x)=\dfrac 1 {\ln'(\exp(x))}=\dfrac{1}{1/\exp(x)}=\exp(x),$$ which is what we wanted to demonstrate.