Derivative

Definition

Let $f$ be defined (and real-valued) on $[a,b]$. For any $x\in [a,b]$ for the quotient

$$\varphi (x)=\frac{f(t)-f(x)}{t-x}(a<t<b,t\ne x)$$

and define

$$f^{\prime }(x)=\underset{t\to x}{\lim }\varphi (t)$$

provided this limit exists. So, $f^{\prime }$ is the derivative of $f$. If $f^{\prime }$ is defined at a point $x$, then $f$ is differentiable at $x$. If this is true $\forall x\in E\subset [a,b]$ then $f$ is differentiable on $E$.

Theorem (Differentiable is Continuous)

Let $f$ be defined on $[a,b]$. If $f$ is differentiable at a point $x\in [a,b]$, then $f$ is continuous at $x$. As $t\to x$, then theorem,

$$f(t)-f(x)=\frac{f(t)-f(x)}{t-x}\cdot (t-x)\to f^{\prime }(x)\cdot 0=0$$

The converse is not true.

Theorem (Derivative Rules)

1. $(f+g{)}^{\prime }(x)=f^{\prime }(x)+g^{\prime }(x)$
2. $(fg{)}^{\prime }(x)=f^{\prime }(x)g(x)+f(x)+g^{\prime }(x)$
3. $(f/g{)}^{\prime }(x)=[g(x)f^{\prime }(x)-g^{\prime }(x)f(x)]/g^2(x)$ where $g(x)\ne 0$.

Indeed, every derivative of a constant is $0$, and $f(x)=x^n$ means $f^{\prime }(x)=n{x}^{n-1}$ by repeating $2$ and $3$ when $n\in \mathbb{Z}$. If $x<0$ then let $n\ne 0$. Thus, every polynomial and rational function is differentiable.

Theorem (Chain Rule)

Suppose $f$ is continuous on $[a,b]$, $f^{\prime }(x)$ exists at some point $x\in [a,b]$, $g$ is defined on an interval $I$ which contains the range of $f$, and $g$ is differentiable at the point $f(x)$. If

$$h(t)=g(f(t))(a\le t\le b)$$

then $h$ is differentiable at $x$, and

$$h^{\prime }(x)=g^{\prime }(f(x))\cdot f^{\prime }(x)$$

Proof: Let $y=f(x)$. By definition of the derivative,

$$\begin{array}{rl}f(t)-f(x)& =(t-x)[f^{\prime }(x)+u(t)]\\ g(s)-g(y)& =(s-y)[g^{\prime }(y)+v(s)]\end{array}$$

where $t\in [a,b]$, $s\in I$ and

• $u(t)\to 0$ as $t\to x$
• $v(s)\to 0$ as $s\to y$

Note that $u,v$ are functions to indicate the “remainder” as $t\to x$.

Let $s=f(t)$. Using the equations above,

$$\begin{array}{rl}h(t)-h(x)& =g(f(t))-g(f(x))\\ & =[f(t)-f(x)]\cdot [g^{\prime }(y)+v(s)]\\ & =(t-x)\cdot [f^{\prime }(t)+u(t)]\cdot [g^{\prime }(y)+v(s)]\end{array}$$

or, if $t\ne x$, then

$$\frac{h(t)-h(x)}{t-x}=[g^{\prime }(y)+v(s)]\cdot [f^{\prime }(x)+u(t)]$$

Letting $t\to x$, we see that $s\to y$, so that RHS tends to $g^{\prime }(y)f^{\prime }(x)$, which gives $h^{\prime }(x)$.

The remainders get smaller.

Definition (Local Maximum/Minimum)

Let $f$ be a real function defined on a Metric Space $X$. We say $f$ has a local maximum at a point $p\in X$ if there exists $\delta >0$ such that $f(q)\le f(p),\forall q\in X$ with $d(p,q)<\delta$.

Local Minima is defined likewise.

Theorem (Local Maxima have Zero-Derivatives)

Let $f$ be defined on $[a,b]$. If $f$ has a local maximum at a point $x\in (a,b)$, and if $f^{\prime }(x)$ exists, then $f^{\prime }(x)=0$.

Proof: Let $\delta >0$ be the radius in which $x$ is the local maximum such that

$$a<x-\delta <x<x+\delta <b$$

If $x-\delta <t<x$, then

$$\begin{array}{r}\frac{f(t)-f(x)}{t-x}\ge 0\end{array}$$

Letting $t\to x$, we see that $f^{\prime }(x)\ge 0$. If $x<t<x+\delta$, then

$$\frac{f(t)-f(x)}{t-x}\le 0$$

and $f^{\prime }(x)\le 0$. Hence $f^{\prime }(x)=0$.

Theorem (Cauchy Mean Value Theorem)

If $f,g$ are continuous real functions on $[a,b]$ which are differentiable in $(a,b)$, then there is a point $x\in (a,b)$ at which

$$[f(b)-f(a)]g^{\prime }(x)=[g(b)-g(a)]f^{\prime }(x)$$

Differentiablility is not required at the endpoints.

Proof: Let

$$h(t)=[f(b)-f(a)]g(t)-[g(b)-g(a)]f(t)(a\le t\le b)$$

Then $h$ is continuous on $[a,b]$, $h$ is differentiable in $(a,b)$, and

$$h(a)=f(b)g(a)-f(a)g(b)=h(b)$$

WTS that $h^{\prime }(x)=0$ for some $x\in (a,b)$. If $h$ is constant, then this is true for every $x\in (a,b)$. If $h(t)>h(a)$ for some $t\in (a,b)$, then let $x\in [a,b]$ that attains the maximum. By theorem, $h^{\prime }(x)=0$. The same argument applies for the minimum.

Theorem (Lagrange Mean Value Theorem)

If $f$ is a real continuous function on $[a,b]$ which is differentiable in $(a,b)$, then there is a point $x\in (a,b)$ at which

$$f(b)-f(a)=(b-a)f^{\prime }(x)$$

Proof: Take $g(x)=x$ in Theorem (Cauchy Mean Value Theorem).

Remark (Rolle’s Theorem)

This is a special case of the MVT when $f(a)=f(b)$. So, $\exists c\in (a,b)$ where $f^{\prime }(c)=0$.

Theorem (Derivative Monotonicity)

Suppose $f$ is differentiable in $(a,b)$.

1. If $\forall x\in (a,b),f^{\prime }(x)\ge 0$ then $f$ is monotonically increasing.
2. If $\forall x\in (a,b),f^{\prime }(x)=0$ then $f$ is constant.
3. If $\forall x\in (a,b),f^{\prime }(x)\le 0$ then $f$ is monotonically decreasing.

Proof: Use Theorem (Lagrange Mean Value Theorem).

Theorem (Darboux’s Theorem)

Suppose $f$ is a real differentiable function on $[a,b]$ and suppose

$$f^{\prime }(a)<\lambda <f^{\prime }(b)$$

Then $\exists x\in (a,b)$ such that $f^{\prime }(x)=\lambda$.

Note this is different from the Intermediate Value Theorem. Note the derivative.

Proof: Put $g(t)=f(t)-\lambda t$.

This is because $g^{\prime }(t)=f^{\prime }(t)-\lambda$ and so we need to find when $g^{\prime }(t)=0$.

Then $g^{\prime }(a)<0$, so that $g(t_1)<g(a)$ for some $t_1\in (a,b)$ and $g^{\prime }(b)>0$, so that $g(t_2)<g(b)$ for some $t_2\in (a,b)$.

See above.

Hence $g$ attains a minimum on $[a,b]$ at some point $a<x<b$ by theorem. By Theorem (Local Maxima have Zero-Derivatives), $g^{\prime }(x)=0$. Hence $f^{\prime }(x)=\lambda$.

Higher Derivatives

If $f$ has a derivative $f^{\prime }$ on an interval, and if $f^{\prime }$ is itself differentiable, we denote the derivative $f^{\prime }$ by $f^{\prime \prime }$ and call $f^{\prime \prime }$ the second derivative of $f$. Continuing, we get

$$f,f^{\prime },f^{\prime \prime },f^{(3)},\cdots ,f^{(n)}$$

each one a derivative of the previous one. In order for $f^{(n)}(x)$ to exist at a point $x$,

1. $f^{(n-1)}(t)$ must exist in a neighborhood of $x$
2. $f^{(n-1)}$ must be differentiable at $x$
3. Since $f^{(n-1)}$ must exist in a neighborhood of $x$, $f^{(n-2)}$ must be differentiable in that neighborhood and so on.