17 Chain Rule, Derivatives of Inverse Functions, Central Theorems (Local Extrema, Rolle's Theorem, Mean Value Theorem)

Lecture from: 24.04.2024 | Video: Video ETHZ

Clicker Question: Functions Satisfying $f^{\prime }=f$

Which differentiable functions $f:\mathbb{R}\to \mathbb{R}$ satisfy the condition $f^{\prime }=f$?

• Is $f(x)=0$ a solution? Yes. If $f(x)=0$ for all $x$, then $f^{\prime }(x)=0$. So, $f^{\prime }(x)=f(x)$ holds.

• Is $f(x)=c\cdot \exp (x)$ for some $c\in \mathbb{R}$ a solution? Yes. Let $f(x)=c\cdot \exp (x)$. Then $f^{\prime }(x)=(c\cdot \exp (x){)}^{\prime }=c\cdot (\exp (x){)}^{\prime }=c\cdot \exp (x)$. Since $c\cdot \exp (x)=f(x)$, we have $f^{\prime }(x)=f(x)$.

• Is $f(x)=\exp (x+d)$ for some $d\in \mathbb{R}$ a solution? Yes. We know $\exp (x+d)=\exp (d)\cdot \exp (x)$. Let $c=\exp (d)$, which is a constant. Then $f(x)=c\cdot \exp (x)$. From the previous point, we know that for such functions, $f^{\prime }=f$. This holds due to the second point above.

Are there any other such functions $f$?

It turns out, no! The functions $f(x)=c\cdot \exp (x)$ for $c\in \mathbb{R}$ are all the real solutions to the “differential equation” $f^{\prime }=f$.

(This topic is explored in much more detail in Analysis II. But a proof will be shown at the end of the lecture note…)

Review: Differentiability

Quick Recap: A function $f:D\to \mathbb{R}$ is differentiable at an accumulation point $x_0\in D$ if the limit $f^{\prime }(x_0)={\lim }_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}$ exists as a finite real number. This $f^{\prime }(x_0)$ is the derivative of $f$ at $x_0$, representing the slope of the tangent line.

Key Corollary: If $f$ is differentiable at $x_0$, then $f$ is continuous at $x_0$.

We also saw a useful reformulation: $f$ is differentiable at $x_0$ if and only if there’s a function $\Phi :D\to \mathbb{R}$ that’s continuous at $x_0$ such that $f(x)=f(x_0)+\Phi (x)(x-x_0)\text{for all }x\in D$ In this case, $\Phi (x_0)=f^{\prime }(x_0)$.

The tangent line at $x_0$ is given by $g(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)$.

Theorem: The Chain Rule

Let $D,E\subseteq \mathbb{R}$. Let $f:D\to E$ and $g:E\to \mathbb{R}$.

Let $x_0\in D$ be an accumulation point of $D$, and let $y_0:=f(x_0)$ be an accumulation point of $E$.

Assume that $f$ is differentiable at $x_0$ and $g$ is differentiable at $y_0=f(x_0)$.

Then the composite function $g\circ f:D\to \mathbb{R}$ is differentiable at $x_0$, and its derivative is: $(g\circ f{)}^{\prime }(x_0)=g^{\prime }(f(x_0))\cdot f^{\prime }(x_0)$

In words: The derivative of a composition is the derivative of the “outer function” (evaluated at the “inner function”) times the derivative of the “inner function.”

Proof (Idea using the $\Phi$-reformulation)

Since $f$ is differentiable at $x_0$, there exists $\Phi :D\to \mathbb{R}$ continuous at $x_0$ with: $f(x)=f(x_0)+\Phi (x)(x-x_0)$ and $\Phi (x_0)=f^{\prime }(x_0)$.

Since $g$ is differentiable at $y_0=f(x_0)$, there exists $\Psi :E\to \mathbb{R}$ continuous at $y_0$ with: $g(y)=g(y_0)+\Psi (y)(y-y_0)$ and $\Psi (y_0)=g^{\prime }(y_0)$.

• Now consider $g(f(x))$.
• Let $y=f(x)$ and $y_0=f(x_0)$.
• $g(f(x))=g(y_0)+\Psi (f(x))(f(x)-y_0)$
• Substitute $f(x)-y_0=f(x)-f(x_0)=\Phi (x)(x-x_0)$:
• $g(f(x))=g(f(x_0))+\Psi (f(x))\Phi (x)(x-x_0)$.

Let $H(x)=\Psi (f(x))\Phi (x)$.

Since $f$ is differentiable at $x_0$, it’s continuous at $x_0$. Since $\Psi$ is continuous at $y_0=f(x_0)$, the composition $\Psi \circ f$ is continuous at $x_0$. Since $\Phi$ is continuous at $x_0$, the product $H(x)=(\Psi \circ f)(x)\cdot \Phi (x)$ is continuous at $x_0$.

So, $g(f(x))=g(f(x_0))+H(x)(x-x_0)$, where $H(x)$ is continuous at $x_0$.

By the reformulation of differentiability (Theorem 4.1.4), $g\circ f$ is differentiable at $x_0$, and $(g\circ f{)}^{\prime }(x_0)=H(x_0)=\Psi (f(x_0))\Phi (x_0)=\Psi (y_0)\Phi (x_0)=g^{\prime }(y_0)f^{\prime }(x_0)=g^{\prime }(f(x_0))f^{\prime }(x_0)$.

Corollary: Derivative of an Inverse Function

Let $f:D\to E$ be bijective, $x_0\in D$ an accumulation point of $D$. Let $f$ be differentiable at $x_0$ with $f^{\prime }(x_0)\ne 0$.

Assume additionally that the inverse function $f^{-1}:E\to D$ is continuous at $y_0=f(x_0)$. Then $y_0$ is an accumulation point of $E$, $f^{-1}$ is differentiable at $y_0$, and:

Remark on the continuity assumption for $f^{-1}$

If $D=I$ is an interval (containing more than one point), and $f:I\to E=f(I)$ is strictly monotonic and continuous, then the assumption that $f^{-1}$ is continuous at $y_0=f(x_0)$ is automatically satisfied due to the theorem on the continuity of inverse functions (from Chapter 3).

Proof (Heuristic, not fully rigorous without establishing differentiability of $f^{-1}$ first)

For all $x\in D$, we have $(f^{-1}\circ f)(x)=f^{-1}(f(x))=x$.

Differentiate both sides with respect to $x$ at the point $x_0$, using the chain rule on the left side (this step assumes $f^{-1}$ is differentiable at $f(x_0)$):

• $(f^{-1}{)}^{\prime }(f(x_0))\cdot f^{\prime }(x_0)=(x{)}^{\prime }=1$.
• Let $y_0=f(x_0)$. Then $(f^{-1}{)}^{\prime }(y_0)\cdot f^{\prime }(x_0)=1$.

Since $f^{\prime }(x_0)\ne 0$, we can divide:

Problem: This argument assumes the differentiability of $f^{-1}$ at $y_0$ beforehand. We actually need to prove this first.

Actual Proof: See script for the rigorous derivation.

Example: Derivative of the Natural Logarithm

1. The natural logarithm $\ln :(0,\infty )\to \mathbb{R}$ is differentiable, and ${\ln }^{\prime }(y)=\frac{1}{y}$ for all $y\in (0,\infty )$.

   Reasoning: $\ln$ is the inverse function of $\exp :\mathbb{R}\to (0,\infty )$. We know ${\exp }^{\prime }(x)=\exp (x)$, which is never $0$ (since $\exp (x)>0$ for all $x\in \mathbb{R}$). Let $y=\exp (x)$, so $x=\ln (y)$.

   Using the corollary for the derivative of the inverse function: ${\ln }^{\prime }(y)=({\exp }^{-1}{)}^{\prime }(y)=\frac{1}{{\exp }^{\prime }(\ln (y))}$.

   Since ${\exp }^{\prime }(\cdot )=\exp (\cdot )$, we have ${\exp }^{\prime }(\ln (y))=\exp (\ln (y))=y$. Therefore, ${\ln }^{\prime }(y)=\frac{1}{y}$.

   Since this holds for every $y\in (0,\infty )$ (which can be written as $\exp (x)$ for some $x\in \mathbb{R}$), the formula is established.

2. Derivative of $x^a$: Let $a\in \mathbb{R}$. Consider the function $(0,\infty )\to (0,\infty )$ given by $x\mapsto x^a=\exp (a\ln (x))$.

   This function is differentiable, and $(x^a{)}^{\prime }=a{x}^{a-1}$ for all $x\in (0,\infty )$.

   Reasoning using the Chain Rule: Let $f(x)=x^a=\exp (a\ln (x))$.

   This is a composition: outer function $u\mapsto \exp (u)$, inner function $x\mapsto a\ln (x)$. $(x^a{)}^{\prime }=(\exp (a\ln (x)){)}^{\prime }={\exp }^{\prime }(a\ln (x))\cdot (a\ln (x){)}^{\prime }$ We know ${\exp }^{\prime }(u)=\exp (u)$.

   The derivative of $a\ln (x)$ with respect to $x$ is $a\cdot {\ln }^{\prime }(x)=a\cdot \frac{1}{x}$.

   So, $(x^a{)}^{\prime }=\exp (a\ln (x))\cdot \left(a\cdot \frac{1}{x}\right)$

   $=x^a\cdot \frac{a}{x}=a\cdot x^a\cdot x^{-1}=a{x}^{a-1}$.

Exercise: Can the statement about $(x^a{)}^{\prime }$ be extended in a suitable sense to the domain $[0,\infty )$? If so, how? (Hint: The answer depends on $a$.)

Central Theorems about the First Derivative

Definition: Local Extrema

Let $f:D\to \mathbb{R}$, $D\subseteq \mathbb{R}$, $x_0\in D$.

1. $f$ has a local maximum at $x_0$ if there exists a $\delta >0$ such that $f(x)\le f(x_0)$ for all $x\in (x_0-\delta ,x_0+\delta )\cap D$.
2. $f$ has a local minimum at $x_0$ if there exists a $\delta >0$ such that $f(x)\ge f(x_0)$ for all $x\in (x_0-\delta ,x_0+\delta )\cap D$.
3. $f$ has a local extremum at $x_0$ if $x_0$ is a local minimum or a local maximum of $f$.

Example

In the image:

• Local Minima at: $x_1,x_3,a$.
• Local Maxima at: $x_0,x_2,b$.

Note: $f(a)$ is a local minimum (assuming $x_0$ is within $(a-\delta ,a+\delta )$ includes only points to the right of $a$ that are in $D$). Similarly for $f(b)$.

It appears from the sketch that $f^{\prime }(x_0)=0$. What about $f^{\prime }(b)$? It looks like $f^{\prime }(b)>0$ in this sketch, so $b$ would not be a point where the derivative is zero, even though it’s a local maximum (because it’s an endpoint).

Theorem: First Derivative Test for Local Extrema (Adapted)

Let $D\subseteq \mathbb{R}$, $x_0\in D$ be an accumulation point of $D$. Let $f:D\to \mathbb{R}$ be differentiable at $x_0$.

1. If $f^{\prime }(x_0)>0$, then there exists a $\delta >0$ such that:

   • $f(x)>f(x_0)$ for all $x\in (x_0,x_0+\delta )\cap D$ (function increases to the right)
   • $f(x)<f(x_0)$ for all $x\in (x_0-\delta ,x_0)\cap D$ (function decreases from the left)

   

2. If $f^{\prime }(x_0)<0$, then there exists a $\delta >0$ such that:

   • $f(x)<f(x_0)$ for all $x\in (x_0,x_0+\delta )\cap D$ (function decreases to the right)
   • $f(x)>f(x_0)$ for all $x\in (x_0-\delta ,x_0)\cap D$ (function increases from the left)

   

3. If $f$ has a local extremum at $x_0$, and $x_0$ is both a left-sided and a right-sided accumulation point of $D$ (i.e., $x_0$ is an interior point of an interval within $D$), then it follows that $f^{\prime }(x_0)=0$. (This excludes, for example, “boundary points” from automatically having $f^{\prime }(x_0)=0$ if they are extrema).

   

Proofs

Part 1

We write $f(x)=f(x_0)+\Phi (x)(x-x_0)$, where $\Phi :D\to \mathbb{R}$ is continuous at $x_0$ and $\Phi (x_0)=f^{\prime }(x_0)$.

Since $f^{\prime }(x_0)>0$, by the continuity of $\Phi$ at $x_0$, there exists a $\delta >0$ such that $\Phi (x)>0$ for all $x\in (x_0-\delta ,x_0+\delta )\cap D$.

Then $f(x)-f(x_0)=\Phi (x)(x-x_0)$.

• If $x\in (x_0,x_0+\delta )\cap D$, then $x-x_0>0$. Since $\Phi (x)>0$, $f(x)-f(x_0)>0$, so $f(x)>f(x_0)$.
• If $x\in (x_0-\delta ,x_0)\cap D$, then $x-x_0<0$. Since $\Phi (x)>0$, $f(x)-f(x_0)<0$, so $f(x)<f(x_0)$.

Part 2

Just like part 1.

Part 3

Suppose $x_0$ is a local extremum and both a left and right-sided accumulation point.

If $f^{\prime }(x_0)>0$, then by part (1), $f(x)<f(x_0)$ for $x$ slightly to the left of $x_0$, and $f(x)>f(x_0)$ for $x$ slightly to the right. This contradicts $x_0$ being a local extremum.

If $f^{\prime }(x_0)<0$, then by part (2), $f(x)>f(x_0)$ for $x$ slightly to the left, and $f(x)<f(x_0)$ for $x$ slightly to the right. This also contradicts $x_0$ being a local extremum.

Therefore, the only possibility left is $f^{\prime }(x_0)=0$.

Theorem: Rolle’s Theorem

Let $a<b$. If $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and if $f(a)=f(b)$, then there exists a $\xi \in (a,b)$ such that $f^{\prime }(\xi )=0$.

Proof (Idea)

Use the Min-Max Theorem (Extreme Value Theorem) and Theorem 4.2.2 (part 3). Since $f$ is continuous on the compact interval $[a,b]$, it attains a global minimum and a global maximum on $[a,b]$.

• Case 1: If the minimum and maximum both occur at the endpoints $a$ and $b$. Since $f(a)=f(b)$, this means the function is constant on $[a,b]$. Then $f^{\prime }(x)=0$ for all $x\in (a,b)$, so any $\xi \in (a,b)$ works.
• Case 2: If at least one extremum (say, a maximum) occurs at an interior point $\xi \in (a,b)$. Then $\xi$ is a local extremum, and since it’s an interior point, $f^{\prime }(\xi )=0$ by Theorem 4.2.2(3).

(Details in script.)

Theorem: Mean Value Theorem (MVT) (Satz von Lagrange)

Let $a<b$. If $f:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a $\xi \in (a,b)$ such that

The term $\frac{f(b)-f(a)}{b-a}$ is the slope of the secant line connecting $(a,f(a))$ and $(b,f(b))$. The MVT says there’s a point $\xi$ where the tangent line is parallel to this secant line.

Proof

Consider the auxiliary function $h(x)=f(x)-g(x)$, where $g(x)$ is the line passing through $(a,f(a))$ and $(b,f(b))$.

The equation of the line $g(x)$ is $g(x)=f(a)+\frac{f(b)-f(a)}{b-a}(x-a)$.

So, its derivative is $g^{\prime }(x)=\frac{f(b)-f(a)}{b-a}$ for all $x\in \mathbb{R}$.

The function $h(x)$ satisfies:

• $h(a)=f(a)-g(a)=f(a)-f(a)=0$.
• $h(b)=f(b)-g(b)=f(b)-\left(f(a)+\frac{f(b)-f(a)}{b-a}(b-a)\right)=f(b)-(f(a)+f(b)-f(a))=0$.

So $h(a)=h(b)=0$.

Also, $h(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$ since $f$ and $g$ are.

By Rolle’s Theorem applied to $h(x)$, there exists a $\xi \in (a,b)$ such that $h^{\prime }(\xi )=0$.

$h^{\prime }(\xi )=f^{\prime }(\xi )-g^{\prime }(\xi )=0⟹f^{\prime }(\xi )=g^{\prime }(\xi )$.

Since $g^{\prime }(\xi )=\frac{f(b)-f(a)}{b-a}$, we have $f^{\prime }(\xi )=\frac{f(b)-f(a)}{b-a}$.

Corollary: What the Mean Value Theorem Tells Us About Functions

The Mean Value Theorem (MVT) is more than just a theoretical curiosity; it’s a powerful tool for understanding the overall behavior of a function based on its derivative. Let $a<b$, and consider functions $f,g:[a,b]\to \mathbb{R}$ that are continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$.

Here are some key consequences:

1. Zero Derivative Implies Constant Function: If the derivative $f^{\prime }(\xi )=0$ for every point $\xi$ in $(a,b)$, then the function $f$ must be constant across the entire interval $[a,b]$.

   Intuition: If the “instantaneous rate of change” is always zero, the function isn’t changing at all.

2. Same Derivative Implies Functions Differ by a Constant: If two functions $f$ and $g$ have the exact same derivative at every point $\xi$ in $(a,b)$ (i.e., $f^{\prime }(\xi )=g^{\prime }(\xi )$), then these functions can only differ by a constant. That is, there’s some $c\in \mathbb{R}$ such that $f(x)=g(x)+c$ for all $x\in [a,b]$.

   Intuition: If they are changing at the same rate everywhere, their difference must remain fixed. Think of two cars moving with the same velocity; the distance between them stays constant.

3. Non-Negative Derivative Implies Increasing Function: If $f^{\prime }(\xi )\ge 0$ for all $\xi \in (a,b)$, then $f$ is (non-strictly) monotonically increasing on $[a,b]$.

   Intuition: If the slope is never negative, the function is always either going up or staying flat.

4. Positive Derivative Implies Strictly Increasing Function: If $f^{\prime }(\xi )>0$ for all $\xi \in (a,b)$, then $f$ is strictly monotonically increasing on $[a,b]$.

   Intuition: If the slope is always positive, the function is always going up.

5. Non-Positive Derivative Implies Decreasing Function: If $f^{\prime }(\xi )\le 0$ for all $\xi \in (a,b)$, then $f$ is (non-strictly) monotonically decreasing on $[a,b]$.

6. Negative Derivative Implies Strictly Decreasing Function: If $f^{\prime }(\xi )<0$ for all $\xi \in (a,b)$, then $f$ is strictly monotonically decreasing on $[a,b]$.

7. Bounded Derivative Implies Lipschitz Continuity: If the derivative is bounded, meaning there’s an $M\ge 0$ such that $|f^{\prime }(\xi )|\le M$ for all $\xi \in (a,b)$, then the function $f$ satisfies a Lipschitz condition on $[a,b]$:

   $$|f(x_1)-f(x_2)|\le M|x_1-x_2|\text{for all }{x}_1,x_2\in [a,b]$$

   Intuition: A bound on the slope means the function can’t change “too quickly.” This is a stronger condition than just continuity.

Remark: This set of results is incredibly useful because it connects local information (the value of the derivative at individual points) to global properties of the function over an entire interval.

Proof (Illustrating for Property 1)

We want to show that if $f^{\prime }(\xi )=0$ for all $\xi \in (a,b)$, then $f$ is constant on $[a,b]$. Pick any two distinct points $x_1,x_2$ in $[a,b]$ with $x_1<x_2$. We want to show $f(x_1)=f(x_2)$.

Apply the Mean Value Theorem to $f$ on the interval $[x_1,x_2]$. (Note that $f$ is continuous on $[x_1,x_2]$ and differentiable on $(x_1,x_2)$ because these conditions hold on the larger interval $[a,b]$).

The MVT guarantees there exists some $\xi \in (x_1,x_2)$ such that

Since $\xi \in (x_1,x_2)\subseteq (a,b)$, our premise states that $f^{\prime }(\xi )=0$. Therefore,

This implies $f(x_2)=f(x_1)$. Since $x_1$ and $x_2$ were arbitrary points in $[a,b]$, the function $f$ must have the same value everywhere on $[a,b]$, meaning $f$ is constant. $\square$

(The proofs for the other properties generally follow a similar pattern: pick two points, apply MVT to the interval between them (or to an auxiliary function like $h(x)=f(x)-g(x)$ for property 2), and use the information about the derivative.)

Application: Solving the Differential Equation $f^{\prime }=f$

We previously asked: which differentiable functions $f:\mathbb{R}\to \mathbb{R}$ satisfy $f^{\prime }(x)=f(x)$ for all $x$? We guessed $f(x)=c{e}^x$. Now we can prove this is the only family of solutions.

Theorem

The only differentiable functions $f:\mathbb{R}\to \mathbb{R}$ satisfying $f^{\prime }(x)=f(x)$ for all $x\in \mathbb{R}$ are of the form $f(x)=c\cdot e^x$, where $c$ is an arbitrary real constant.

Proof

Suppose $f(x)$ is a differentiable function such that $f^{\prime }(x)=f(x)$. Consider an auxiliary function $g(x)=e^{-x}f(x)$.

Let’s find the derivative of $g(x)$ using the product rule:

We know $(e^{-x}{)}^{\prime }=-e^{-x}$ (using the chain rule, or by knowing the derivative of $e^{kx}$ is $k{e}^{kx}$).

So,

Now, we use the given condition that $f^{\prime }(x)=f(x)$:

Since $g^{\prime }(x)=0$ for all $x\in \mathbb{R}$, by Corollary 4.2.5 (part 1), the function $g(x)$ must be a constant. Let’s call this constant $c$.

So, $g(x)=c$, which means

To solve for $f(x)$, we can multiply both sides by $e^x$ (which is always non-zero):

This shows that any function satisfying $f^{\prime }=f$ must be of this form. We already verified earlier that any function of this form indeed satisfies $f^{\prime }=f$.

Continue here: 18 Derivatives of Inverse Trig Functions, L’Hopital’s Rule, Convex Functions