Derivative of an Inverse Function
Let f(x) be a continuous function that is strictly increasing (or strictly decreasing) on an interval [a,b]. If f is differentiable at a point x in (a,b) and f'(x)≠0, then its inverse function f-1 is also differentiable at the corresponding point y=f(x). The derivative of the inverse function is given by $$ D f^{-1}(y)=\frac{1}{f'(x)} = \frac{1}{f'(f^{-1}(y))} $$
This result allows you to compute the derivative of an inverse function without differentiating the inverse directly.
A Practical Example
Consider the function
\[ y=x^3 \]
Its inverse function is
\[ x=\sqrt[3]{y} \]
The derivative of the original function is
\[ f'(x)=3x^2 \]
Using the inverse function derivative formula, we obtain
\[ (f^{-1})'(y)=\frac{1}{f'(x)} \]
\[ (f^{-1})'(y)=\frac{1}{3x^2} \]
The result is still expressed in terms of x. To rewrite it in terms of y, use the inverse function:
\[ x=\sqrt[3]{y} \]
Substituting this expression into the formula gives
\[ (f^{-1})'(y) = \frac{1}{3(\sqrt[3]{y})^2} \]
Therefore,
\[ (f^{-1})'(y) = \frac{1}{3y^{2/3}} \]
or, equivalently,
\[ (f^{-1})'(y) = \frac{1}{3\sqrt[3]{y^2}} \]
Note. You can verify the result by differentiating the inverse function directly. Since $$ x=\sqrt[3]{y}=y^{1/3} $$ we have $$ (f^{-1})'(y)=D[y^{1/3}] $$ $$ (f^{-1})'(y)=\frac{1}{3}y^{(1/3-1)} $$ $$ (f^{-1})'(y)=\frac{1}{3}y^{(1-3)/3} $$ $$ (f^{-1})'(y)=\frac{1}{3}y^{-2/3} $$ $$ (f^{-1})'(y)=\frac{1}{3y^{2/3}} $$ $$ (f^{-1})'(y)=\frac{1}{3\sqrt[3]{y^2}} $$ As expected, this is exactly the same result.
Example 2
Now consider the function
$$ y=f(x)=x^2 $$
This function is strictly increasing for x>0.
Why? For any two positive values x1 and x2, the following implication holds: $$ x_1 < x_2 \Rightarrow f(x_1) < f(x_2) $$
Since the function is one-to-one on the interval x>0, it has an inverse function.
The inverse function is
$$ x=f^{-1}(y)=\sqrt{y} $$
How is the inverse obtained? Starting from $$ y=x^2 $$ take the square root of both sides: $$ \sqrt{y}=\sqrt{x^2} $$ $$ \sqrt{y}=x $$ Since we are restricting the domain to x>0, the inverse function is $$ x=f^{-1}(y)=\sqrt{y} $$ For example, if x=3, then $$ y=f(x)=3^2=9 $$ and the inverse function returns $$ x=f^{-1}(9)=\sqrt{9}=3 $$
The derivative of the original function is
$$ D[f(x)]=D[x^2]=2x $$
Therefore, the derivative of the inverse function is
$$ D[f^{-1}(y)]=D[\sqrt{y}]=\frac{1}{2\sqrt{y}} $$
The same result follows from the inverse function derivative formula:
$$ D f^{-1}(y)=\frac{1}{f'(x)}=\frac{1}{f'(f^{-1}(y))} $$
Since \(x=f^{-1}(y)=\sqrt{y}\),
$$ D f^{-1}(y)=\frac{1}{2x}=\frac{1}{2\sqrt{y}} $$
Both approaches lead to the same answer.
Derivatives of Inverse Trigonometric Functions
The inverse function differentiation formula is especially useful when finding the derivatives of inverse trigonometric functions such as arcsine, arccosine, and arctangent.
Derivative of the Arcsine Function
The arcsine function is defined for $ x \in [-1,1] $.
$$ y = \arcsin x $$
Its inverse relation is
$$ x = \sin y $$
where the variable $ y $ is restricted to the interval
$$ y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $$
To find the derivative of the arcsine function with respect to $ x $, we apply the inverse function differentiation formula.
$$ \frac{ d \ \arcsin x }{dx} = \frac{1}{\frac{ d \ \sin y }{dy}} $$
Since
$$ \frac{ d \ \sin y }{dy} = \cos y $$
we obtain
$$ \frac{ d \ \arcsin x }{dx} = \frac{1}{\cos y} $$
The result is still expressed in terms of $ y $, so the next step is to rewrite it using the variable $ x $.
Note. Using the Pythagorean trigonometric identity, $$ \cos^2 y + \sin^2 y = 1 $$ we can write $$ \cos y = \sqrt{1-\sin^2 y} $$ because \( y \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \), where the cosine function is always non-negative. Therefore, the positive square root must be taken.
Substituting this expression into the derivative gives
$$ \frac{ d \ \arcsin x }{dx} = \frac{1}{\sqrt{1-\sin^2 y}} $$
Since $ x = \sin y $, we can replace $ \sin y $ with $ x $.
$$ \frac{ d \ \arcsin x }{dx} = \frac{1}{\sqrt{1-x^2}} $$
This is the derivative of the arcsine function.
$$ \frac{ d \ \arcsin x }{dx} = \frac{1}{\sqrt{1-x^2}} $$
Other Inverse Trigonometric Functions
The derivatives of the other inverse trigonometric functions can be obtained using the same approach.
\[ \frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1-x^2}} \]
\[ \frac{d}{dx}(\arctan x) = \frac{1}{1+x^2} \]
\[ \frac{d}{dx}(\operatorname{arccot} x) = -\frac{1}{1+x^2} \]
Proof
Let
$$ x=f^{-1}(y) $$
By definition, the derivative of the inverse function is the limit of its difference quotient:
$$ D[f^{-1}(y)] = \lim_{h \rightarrow 0} \frac{f^{-1}(y+h)-f^{-1}(y)}{h} $$
Define
$$ \Delta x=f^{-1}(y+h)-f^{-1}(y) $$
Then
$$ D[f^{-1}(y)] = \lim_{h \rightarrow 0} \frac{\Delta x}{h} $$
Since
$$ h=f(x+\Delta x)-f(x) $$
we can rewrite the derivative as
$$ D[f^{-1}(y)] = \lim_{h \rightarrow 0} \frac{\Delta x}{f(x+\Delta x)-f(x)} $$
Rearranging the expression gives
$$ D[f^{-1}(y)] = \lim_{h \rightarrow 0}\frac{1}{\frac{f(x+\Delta x)-f(x)}{\Delta x}} $$
As h approaches zero, Δx also approaches zero because
$$ \Delta x=f^{-1}(y+h)-f^{-1}(y) $$
Therefore,
$$ D[f^{-1}(y)] = \lim_{\Delta x \rightarrow 0}\frac{1}{\frac{f(x+\Delta x)-f(x)}{\Delta x}} $$
The denominator is the difference quotient of f(x), whose limit is f'(x).
Hence,
$$ D[f^{-1}(y)] = \frac{1}{f'(x)} $$
This proves the formula for the derivative of an inverse function.
Alternative Proof
Consider the function
$$ y=f(x) $$
and its inverse
$$ x=f^{-1}(y) $$
Differentiate both sides with respect to x:
$$ D[x]=D[f^{-1}(y)] $$
Since
$$ D[x]=1 $$
we obtain
$$ 1=D[f^{-1}(y)] $$
The right-hand side is the derivative of a composite function. Applying the chain rule gives
$$ D[f^{-1}(y)]=(f^{-1})'(y)\cdot y' $$
Therefore,
$$ 1=(f^{-1})'(y)\cdot y' $$
Since
$$ y'=f'(x) $$
it follows that
$$ 1=(f^{-1})'(y)\cdot f'(x) $$
and therefore
$$ (f^{-1})'(y)=\frac{1}{f'(x)} $$
Corollary
If f'(x0)=0, then the inverse function is not differentiable at the corresponding point y0=f(x0), assuming that f remains one-to-one near x0.
$$ f'(x_0)=0 \Rightarrow (f^{-1})'(y_0) \text{ does not exist} $$
Geometrically, a horizontal tangent on the graph of f becomes a vertical tangent on the graph of its inverse.
Conversely, if the graph of f has a vertical tangent at x0, then the graph of the inverse function has a horizontal tangent at y0.
$$ f'(x_0) \text{ does not exist and the tangent is vertical} \Rightarrow (f^{-1})'(y_0)=0 $$
In other words, taking the inverse of a function swaps horizontal and vertical tangents.
