Real functions/Differentiability/Rules/Section

These rules are called sum rule, product rule, quotient rule. The following statement is called chain rule.

An illustration of the product rule: the increment of the area is about the seize of the sum of the two products of the side length and the increment of the other side length. For the infinitesimal increment, the product of the two increments is irrelevant.

Lemma

Let ${}D\subseteq \mathbb {R}$ be a subset, ${}a\in D$ a point, and

$f,g\colon D\longrightarrow \mathbb {R}$

functions which are differentiable

in ${}a$. Then the following rules for differentiability holds.

1. The sum ${}f+g$ is differentiable in ${}a$, with

   ${}(f+g)'(a)=f'(a)+g'(a)\,.$

2. The product ${}f\cdot g$ is differentiable in ${}a$, with

   ${}(f\cdot g)'(a)=f'(a)g(a)+f(a)g'(a)\,.$

3. For ${}c\in \mathbb {R}$, also ${}cf$ is differentiable in ${}a$, with

   ${}(cf)'(a)=cf'(a)\,.$

4. If ${}g$ has no zero in ${}a$, then ${}1/g$ is differentiable in ${}a$, with

   ${}{\left({\frac {1}{g}}\right)}'(a)={\frac {-g'(a)}{(g(a))^{2}}}\,.$

5. If ${}g$ has no zero in ${}a$, then ${}f/g$ is differentiable in ${}a$, with

   ${}{\left({\frac {f}{g}}\right)}'(a)={\frac {f'(a)g(a)-f(a)g'(a)}{(g(a))^{2}}}\,.$

Proof

(1). We write ${}f$ and ${}g$ respectively with the objects which were formulated in fact, that is

${}f(x)=f(a)+s(x-a)+r(x)(x-a)\,$

and

${}g(x)=g(a)+{\tilde {s}}(x-a)+{\tilde {r}}(x)(x-a)\,.$

Summing up yields

${}f(x)+g(x)=f(a)+g(a)+(s+{\tilde {s}})(x-a)+(r+{\tilde {r}})(x)(x-a)\,.$

Here, the sum ${}r+{\tilde {r}}$ is again continuous in ${}a$, with value ${}0$.
(2). We start again with

${}f(x)=f(a)+s(x-a)+r(x)(x-a)\,$

and

${}g(x)=g(a)+{\tilde {s}}(x-a)+{\tilde {r}}(x)(x-a)\,,$

and multiply both equations. This yields

${}{\begin{aligned}f(x)g(x)&=(f(a)+s(x-a)+r(x)(x-a))(g(a)+{\tilde {s}}(x-a)+{\tilde {r}}(x)(x-a))\\&=f(a)g(a)+(sg(a)+{\tilde {s}}f(a))(x-a)\\&\,\,\,\,\,+(f(a){\tilde {r}}(x)+g(a)r(x)+s{\tilde {s}}(x-a)+s{\tilde {r}}(x)(x-a)+{\tilde {s}}r(x)(x-a)+r(x){\tilde {r}}(x)(x-a))(x-a).\end{aligned}}$

Due to fact for limits, the expression consisting of the last six summands is a continuous function, with value ${}0$ for ${}x=a$.
(3) follows from (2), since a constant function is differentiable with derivative ${}0$.
(4). We have

${}{\frac {{\frac {1}{g(x)}}-{\frac {1}{g(a)}}}{x-a}}={\frac {-1}{g(a)g(x)}}\cdot {\frac {g(x)-g(a)}{x-a}}\,.$

Since ${}g$ is continuous in ${}a$, due to fact, the left-hand factor converges for ${}x\rightarrow a$ to ${}-{\frac {1}{g(a)^{2}}}$, and because of the differentiability of ${}g$ in ${}a$, the right-hand factor converges to ${}g'(a)$.
(5) follows from (2) and (4).

$\Box$

Theorem

Let ${}D,E\subseteq \mathbb {R}$ denote subsets, and let

$f\colon D\longrightarrow \mathbb {R}$

and

$g\colon E\longrightarrow \mathbb {R}$

be functions with ${}f(D)\subseteq E$. Suppose that ${}f$ is differentiable in ${}a$ and that ${}g$ is differentiable in ${}b:=f(a)$. Then also the composition

$g\circ f\colon D\longrightarrow \mathbb {R}$

is differentiable in ${}a$, and its derivative is

${}(g\circ f)'(a)=g'(f(a))\cdot f'(a)\,.$

Proof

Due to fact, one can write

${}f(x)=f(a)+f'(a)(x-a)+r(x)(x-a)\,$

and

${}g(y)=g(b)+g'(b)(y-b)+s(y)(y-b)\,.$

Therefore,

${}{\begin{aligned}g(f(x))&=g(f(a))+g'(f(a))(f(x)-f(a))+s(f(x))(f(x)-f(a))\\&=g(f(a))+g'(f(a)){\left(f'(a)(x-a)+r(x)(x-a)\right)}+s(f(x)){\left(f'(a)(x-a)+r(x)(x-a)\right)}\\&=g(f(a))+g'(f(a))f'(a)(x-a)+{\left(g'(f(a))r(x)+s(f(x))(f'(a)+r(x))\right)}(x-a).\end{aligned}}$

The remainder function

${}t(x):=g'(f(a))r(x)+s(f(x))(f'(a)+r(x))\,$

is continuous in ${}a$ with value ${}0$.

$\Box$

An illustration for the derivative of the inverse function. The graph of the inverse function is the reflection of the graph at the diagonal, and the tangent behaves accordingly.

Theorem

Let ${}D,E\subseteq \mathbb {R}$ denote intervals, and let

$f\colon D\longrightarrow E\subseteq \mathbb {R}$

be a bijective continuous function, with the inverse function

$f^{-1}\colon E\longrightarrow D.$

Suppose that ${}f$ is

differentiable in ${}a\in D$ with ${}f'(a)\neq 0$. Then also the inverse function ${}f^{-1}$ is differentiable in ${}b:=f(a)$, and

${}(f^{-1})'(b)={\frac {1}{f'(f^{-1}(b))}}={\frac {1}{f'(a)}}\,$

holds.

Proof

We consider the difference quotient

${}{\frac {f^{-1}(y)-f^{-1}(b)}{y-b}}={\frac {f^{-1}(y)-a}{y-b}}\,,$

and have to show that the limit for ${}y\rightarrow b$ exists, and obtains the value claimed. For this, let ${}{\left(y_{n}\right)}_{n\in \mathbb {N} }$ denote a sequence in ${}E\setminus \{b\}$, converging to ${}b$. Because of fact, the function ${}f^{-1}$ is continuous. Therefore, also the sequence with the members ${}x_{n}:=f^{-1}(y_{n})$ converges to ${}a$. Because of bijectivity, ${}x_{n}\neq a$ for all ${}n$. Thus

${}\lim _{n\rightarrow \infty }{\frac {f^{-1}(y_{n})-a}{y_{n}-b}}=\lim _{n\rightarrow \infty }{\frac {x_{n}-a}{f(x_{n})-f(a)}}={\left(\lim _{n\rightarrow \infty }{\frac {f(x_{n})-f(a)}{x_{n}-a}}\right)}^{-1}\,,$

where the right-hand side exists, due to the condition, and the second equation follows from fact  (5).

$\Box$

Example

The function

$f^{-1}\colon \mathbb {R} _{+}\longrightarrow \mathbb {R} _{+},x\longmapsto {\sqrt {x}},$

is the inverse function of the function ${}f$, given by ${}f(x)=x^{2}$ (restricted to ${}\mathbb {R} _{+}$). The derivative of ${}f$ in a point ${}a$ is ${}f'(a)=2a$. Due to fact, for ${}b\in \mathbb {R} _{+}$, the relation

${}{\left(f^{-1}\right)}'(b)={\frac {1}{f'(f^{-1}(b))}}={\frac {1}{2{\sqrt {b}}}}={\frac {1}{2}}b^{-{\frac {1}{2}}}\,$

holds. In the zero point, however, ${}f^{-1}$ is not differentiable.

Example

The function

$f^{-1}\colon \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto x^{\frac {1}{3}},$

is the inverse function of the function ${}f$, given by ${}f(x)=x^{3}$. The derivative of ${}f$ in ${}a$ is ${}f'(a)=3a^{2}$, which is different from ${}0$ for ${}a\neq 0$. Due to fact, we have for ${}b\neq 0$ the relation

${}{\left(f^{-1}\right)}'(b)={\frac {1}{f'(f^{-1}(b))}}={\frac {1}{3{\left(b^{\frac {1}{3}}\right)}^{2}}}={\frac {1}{3}}b^{-{\frac {2}{3}}}\,.$

In the zero point, however, ${}f^{-1}$ is not differentiable.