In the following article we will investigate the conditions under which the inverse function of a bijective function is differentiable at one point. We will also derive a formula with which we can explicitly determine the derivative of the inverse function. The practical thing about this formula is that it allows us to determine the derivative at certain points, even if we do not know the inverse function explicitly or it is insanely difficult.
Let us first consider a linear function as an example. For this it is very easy to determine the derivative of the inverse function. Non-constant linear functions are bijective and therefore invertible on . In this case we can calculate the inverse function explicitly and differentiate it. Concretely we choose with . The inverse function is
is differentiable on and for all .
Let us next consider the function . Here we have to be careful, because it is not injective on all of and therefore not invertible. But if we restrict the domain of definition to , then is bijective. The inverse function is the square root function
For differentiability we have to consider another thing: is not differentiable at . We can show this by examining the differential quotient. Or we consider the following:
Since the root function is the inverse function of the square function , there is . At zero there is thus in particular
If now was differentiable at 0, then the chain rule would yield
So cannot be differentiable at 0. However, on , the function is differentiable, and there is
This example shows that in case , it may happen that is not differentiable although is differentiable everywhere.
In the two examples it was relatively easy to determine the derivative of the inverse function directly (it was a polynomial). But what about more complicated functions, for example as an inverse function of ? Here we cannot simply calculate the derivative of the inverse function, if only derivatives of exponentials and polynomials are known. It may even occur that a bijective function cannot be inverted explicitly. In these cases it would be good to have a general formula with which we can determine the derivative of from the derivative of . If we look again at the derivative from the second example, we may see the following:
Since there is for all and for all . In the first example (straight lines), there is also
Can this be chance? Actually, it's not: the formula is valid for a general. Consider being differentiable at and being differentiable at . By definition of the inverse function,
for all . Now we take the derivative and obtain by the chain rule:
Here we have used that in and in are differentiable. Now we divide on both sides by (note: this only possible if the expression is not equal to zero), and get
So the formula also holds in general under certain conditions. Now the question is, under which conditions at the derivative of exists.
- On the one hand the must exist. This is exactly the case if is bijective, which is exactly the case if is surjective and strictly monotonous.
- As we have seen above, must be differentiable in the point with .
- We will see that we need one more condition, namely that is continuous in . If the domain of definition of is an interval, then this is always fulfilled according to the theorem about continuity of the inverse function.
These are the conditions necessary for our formula to hold. Let's put it into a theorem:
Theorem: derivative of the inverse function [Bearbeiten]
Theorem (Derivative of the inverse function)
Let and be an interval. further, let be a surjective and strictly monotonous function, which is differentiable in where . Then, has an inverse function , which is differentiable at and there is:
- The surjectivity of is equivalent to .
- If is differentiable on all of , then according to the monotony criterion the strict monotony can be seen most easily by or .
- As we have seen above with the derivative of the square root function in , the condition must not be omitted under any circumstances. Otherwise, it produces "infinite derivatives", which are not well-defined!
- The theorem also holds if is not an interval. But then it must be demanded additionally that in is continuous. Furthermore, and must be accumulation points of and respectively.
- If is additionally continuous, then by continuity of the inverse function it follows that is an interval.
Summary of proof (Derivative of the inverse function)
First of all we justify that exists. Then we conclude by the theorem about the continuity of the inverse function that is continuous. We show that the differential quotient exists and has the value . That is, that for every sequence with there is .
Proof (Derivative of the inverse function)
is surjective and strictly monotonous, i.e. bijective. So the inverse function exists. Since we have assumed that is an interval, the theorem about the continuity of the inverse function implies that is continuous on . There is thus with . Let now be a sequence in with , then there is
Hence, is differentiable in and there is .
Alternative proof (Derivative of the inverse function)
Another way of proof is given by an equivalent characterization of the derivative: is differentiable in if and only if there is a function continuous at with
If this is the case, then . Since by assumption, and is strictly monotonous, follows for all . If we now set and , the above equation is
This is now equivalent to
Since and are continuous at we also get continuity of at . If we now use again the equivalent characterization of continuity, it follows from the last equation that is differentiable in with
Memory rule and visualization[Bearbeiten]
Using Leibniz's notation for the derivative, the formula of the derivative of the inverse function can be illustrated by a simple fraction-swap trick: For and there is
We can also visualize the formula graphically: If the function is differentiable at , then corresponds to the slope of the tangent to the graph in . Hence,
We now obtain the graph of the inverse function in two steps:
- First we have to rotate the graph of by (clockwise or counter-clockwise). The resulting graph has the slope at the point , because the tangent at this point is perpendicular to the original tangent.
- Then we have to mirror the graph (horizontally or vertically). The sign of the tangent gradient is reversed.
Altogether we get
Extension to the whole domain[Bearbeiten]
The converse of the theorem also holds:
Theorem (Converse of the theorem about inverse function derivative)
Let and be an interval. Further, let be a surjective, strictly monotonous function, which is differentiable at . If further, the inverse function is differentiable at , then there is: and
Proof (Converse of the theorem about inverse function derivative)
The proof works with the trick from the introduction. For all we have
Under the above conditions, the left-hand side is differentiable at (chain rule) with
Because 0 has no divisor (other than 0) in , there must be and we get
Let us now additionally demand in the original theorem that is differentiable on all of with . Then we can determine the derivative function of on all of :
Theorem (Derivative of the inverse function)
Let and be an interval. Further, let be a surjective, differentiable, strictly monotonous function with for all . Then has a differentiable inverse function, whose derivative is given by:
Example (linear functions)
Let , and
a linear function. Then is surjective and strictly monotonously increasing, if , and strictly monotonously decreasing, if . Furthermore, is differentiable on all of with derivative . According to the theorem about the derivative of the inverse function there is thus for all
We could also have calculated this directly, as above.
Example (Root functions)
Then is differentiable and has the derivative . So it is monotonously increasing. Furthermore, is surjective. The inverse function is the -th root function
For every our theorem now yields
If is odd, then the formula holds even for all .
Example (Logarithmic functions)
Let us look at the exponential function
We have learned that . So the function is differentiable, and because of strictly monotonously increasing. Furthermore, is surjective. The inverse function is the (natural) logarithm function
Our theorem now implies for :
Exercise (Derivative of the inverse function)
Prove that die function
has a differentiable inverse function . Determine the domain of definition of and calculate .
Solution (Derivative of the inverse function)
We have to check all conditions of the theorem about the derivative of the inverse function, one after the other.
Proof step: is surjective
is continuous on as it is a composition of continuous functions. Further, there is
By the intermediate value theorem, for every there is an with . So is surjective.
Proof step: is strictly monotonous
is differentiable on as it is a composition of continuous functions and
for all . According to the monotony criterion, is strictly monotonically decreasing, and therefore injective on .
So is bijective, and thus has a in inverse function . The domain of definition corresponds to the range of values of .
Proof step: is differentiable on and for all
Differentiability was proven in step. As there is also for all .
According to the theorem about the derivative of the inverse function, it is differentiable on all of .
Proof step: Computation of von
There is . Hence , and with the formula for the derivative of the inverse function there is
Exercise (Second derivative of the inverse function)
Let with , and be a twice differentiable bijective function with . Show that the inverse function is twice differentiable, as well and express the second derivative of at the position by derivatives of at a suitable position.
As application: Compute for the polynomial the derivatives and .
Solution (Second derivative of the inverse function)
Proof step: First derivative of
is an interval and is bijective. Because of there is an with . Since there is . According to the theorem about the derivative of the inverse function , is differentiable in with
Proof step: Second derivative of
is twice differentiable. This means that is differentiable. According to the quotient and chain rule is therefore also differentiable at and there is
Proof step: Computing the derivatives and
is differentiable on with . So is strictly monotonously increasing and therefore injective. Because , according to the intermediate value theorem is also surjective. So, ion total bijective. With the derivative theorem of the inverse function implies
Further, is twice differentiable with . With the formula proven in step 2, there is hence