Derivative of an Inverse Function

Let $f(x)$ be a function that is continuous and strictly increasing (or strictly decreasing) on an interval $[a, b]$. If $f$ is differentiable at a point $x$ in $(a, b)$ and $f'(x) \ne 0$, then its inverse function $f^{-1}$ is also differentiable at the point $y = f(x)$, and its derivative is given by: $$ D f^{-1}(y) = \frac{1}{f'(x)} = \frac{1}{f'\bigl(f^{-1}(y)\bigr)}. $$

A Practical Example

Consider the function:

$$ y = f(x) = x^2 $$

This function is strictly increasing for $x > 0$.

Note. A function is strictly increasing if, for any two values $x_1$ and $x_2$ both greater than zero, the following relationship holds: $$ x_1 < x_2 \ \Rightarrow \ f(x_1) < f(x_2). $$

The inverse function of $f(x)$ is:

$$ x = f^{-1}(y) = \sqrt{y}. $$

Explanation. Given the function $$ y = x^2, $$ to express $x$ in terms of $y$, we take the square root of both sides: $$ \sqrt{y} = \sqrt{x^2} $$ $$ \sqrt{y} = x $$ This yields the inverse function: $x = f^{-1}(y)$.

Example. If $x = 3$, then $$ y = f(x) = x^2 = 3^2 = 9. $$ For $y = 9$, the inverse function returns $$ x = f^{-1}(y) = \sqrt{y} = \sqrt{9} = 3. $$

Since $f(x)$ is differentiable for $x > 0$ and strictly increasing, its derivative is:

$$ D[f(x)] = D[x^2] = 2x. $$

The derivative of the inverse function likewise exists for $x > 0$:

$$ D[f^{-1}(y)] = D[\sqrt{y}] = \frac{1}{2 \sqrt{y}}. $$

Applying the inverse function derivative theorem, we find the same result:

$$ D f^{-1}(y) = \frac{1}{f'(x)} = \frac{1}{f'\bigl(f^{-1}(y)\bigr)}. $$

$$ D f^{-1}(y) = \frac{1}{D[x^2]} = \frac{1}{2x} = \frac{1}{2 \sqrt{y}}. $$

Thus, both methods yield consistent results.

Proof

Given a function $f(x)$, its inverse function satisfies:

$$ x = f^{-1}(y). $$

The derivative of the inverse function $f^{-1}$ is defined as the limit of its difference quotient as $h$ approaches zero:

$$ D[f^{-1}(y)] = \lim_{h \to 0} \frac{f^{-1}(y + h) - f^{-1}(y)}{h}. $$

Let:

$$ \Delta x = f^{-1}(y + h) - f^{-1}(y). $$

Then the derivative becomes:

$$ D[f^{-1}(y)] = \lim_{h \to 0} \frac{\Delta x}{h}. $$

Since:

$$ h = f(x + \Delta x) - f(x), $$

we can rewrite the limit as:

$$ D[f^{-1}(y)] = \lim_{h \to 0} \frac{\Delta x}{f(x + \Delta x) - f(x)}. $$

This is equivalent to:

$$ D[f^{-1}(y)] = \lim_{h \to 0} \frac{1}{\frac{f(x + \Delta x) - f(x)}{\Delta x}}. $$

As $h \to 0$, we also have $\Delta x \to 0$ because:

$$ \Delta x = f^{-1}(y + h) - f^{-1}(y). $$

Thus, we can express the limit entirely in terms of $\Delta x$:

$$ D[f^{-1}(y)] = \lim_{\Delta x \to 0} \frac{1}{\frac{f(x + \Delta x) - f(x)}{\Delta x}}. $$

The denominator represents the difference quotient for $f(x)$, which approaches the derivative $f'(x)$ as $\Delta x \to 0$:

$$ D[f^{-1}(y)] = \frac{1}{f'(x)}. $$

Hence:

$$ D[f^{-1}(y)] = \frac{1}{f'(x)}. $$

This completes the proof of the rule for differentiating inverse functions.

Corollary

If the derivative $f'(x)$ equals zero at a point $x_0$, then the inverse function $f^{-1}$ is not differentiable at the corresponding point $y_0 = f(x_0)$, and vice versa:

$$ \frac{df(x_0)}{dx} = 0 \ \Leftrightarrow \ \nexists \ \frac{d f^{-1}(y_0)}{dy}. $$

Conversely, if the derivative $f'(x)$ does not exist at $x_0$, then the derivative of the inverse function $f^{-1}$ is zero at $y_0 = f(x_0)$, and vice versa:

$$ \nexists \ \frac{df(x_0)}{dx} \ \Leftrightarrow \ \frac{d f^{-1}(y_0)}{dy} = 0. $$

And so on.