Differentiation of inverse functions

Let $y$ be given as a function of $x$ by means of the relation $y = f(x)$.

It is usually possible in the case of functions considered in this book to solve this equation for $x$, giving

$$x = \phi (y);$$

that is, to consider $y$ as the independent and $x$ as the dependent variable. In that case $f(x)$ and $\phi (y)$ are said to be inverse functions. When we wish to distinguish between the two it is customary to call the first one given the direct function and the second one the inverse function. Thus, in the examples which follow, if the second members in the first column are taken as the direct functions, then the corresponding members in the second column will be respectively their inverse functions.

Example 5.12.1  

• $y=x^2 + 1$, $x = \pm \sqrt{y - 1}$.
• $y = a^x$, $x = \log_a y$.

• $y = \sin\ x$, $x = \arcsin\ y$.

The plot of the inverse function $\phi (y)$ is related to the plot of the function $f(x)$ in a simple manner. The plot of $f(x)$ over an interval $(a,b)$ in which $f$ is increasing is the same as the plot of $\phi (y)$ over $(f(a),f(b))$.

Example 5.12.2   If $f(x)=x^2$, for $x > 0$, and $\phi(y)=\sqrt{y}$, then the graphs are

Figure 5.1: The function $f(x)=x^2$.

Now flip this graph about the $45^o$ line:

Figure 5.2: The function $\phi (y)=f^{-1}(y)=\sqrt {y}$.

The graph of inverse trig functions, for example, $\tan(x)$ and $\arctan(x)$, are related in the same way.

Let us now differentiate the inverse functions

$$y = f(x)\ \ {\rm and}\ \ x = \phi(y)$$

simultaneously by the General Rule.

• FIRST STEP. $y + \Delta y = f(x + \Delta x)$, $x + \Delta x =\phi(y + \Delta y)$
• SECOND STEP.
  $$\begin{array}{rlcrl} y + \Delta y & = f(x + \Delta x), & \qq... ...quad & \Delta x &= \phi(y + \Delta y) - \phi (y)\\ \end{array}$$

• THIRD STEP.
  $$\frac{\Delta y}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta ... ...uad \frac{\Delta x}{\Delta y} = \frac{\phi(y + \Delta y) - \phi(y)}{\Delta y}.$$
  Taking the product of the left-hand forms of these ratios, we get $\frac{\Delta y}{\Delta x} \cdot \frac{\Delta x}{\Delta y} = 1$, or, $\frac{\Delta y}{\Delta x} = \frac{1}{\frac{\Delta x}{\Delta y}}$.

• FOURTH STEP. Passing to the limit,

  $$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}},$$ (5.2)

  
  
  or,
  $$f'(x) = \frac{1}{\phi'(y)}.$$

The derivative of the inverse function is equal to the reciprocal of the derivative of the direct function.