7.6. Calculation Rules for Differentiable Functions

Proposition (derivation rules, local version):  Let ${\displaystyle a\in A\cap B}$ be an accumulation point of ${\displaystyle A\cap B}$. If  ${\displaystyle f:A\to ℝ}$ and ${\displaystyle g:B\to ℝ}$ are differentiable at a, so are  ${\displaystyle f+g,f-g,f\cdot g}$ and - in case ${\displaystyle g(a)\ne 0}$ - ${\displaystyle \frac{f}{g}}$ as well. In addition we have the

Proof:  All the rules follow directly from the limit theorems [6.9.5] - [6.9.8] using the respective calculation of the difference quotient function in [7.2.7] - [7.2.10]. Note that, as  f and g are differentiable at a, they are as well continuous thus their own value at a coincides with their limit.

With the quotient rule it is important to know that a is still an accumulation point of the domain of the quotient function (see the proof of [6.9.8] for details).

Example:  As ${\displaystyle {\mathrm{X}}^n}$, ${\displaystyle \sqrt{\mathrm{X}}}$ as well as sin and exp are differentiable functions due to [7.3.3/5] and [7.5.8/9] the derivation rules guarantee the following functions to be differentiable at the respectively chosen points. Their derivate numbers are calculated according to 1. to 7.

• ${\displaystyle ({\mathrm{X}}^2+\sin {)}^{\prime }(\mathrm{\pi })=({\mathrm{X}}^2{)}^{\prime }(\mathrm{\pi })+{\sin }^{\prime }(\mathrm{\pi })=2\mathrm{\pi }+\cos \mathrm{\pi }=2\mathrm{\pi }-1}$

• ${\displaystyle ({\mathrm{X}}^3\sqrt{\mathrm{X}}{)}^{\prime }(4)=({\mathrm{X}}^3{)}^{\prime }(4)\cdot \sqrt{\mathrm{X}}(4)+{\mathrm{X}}^3(4)\cdot {\sqrt{\mathrm{X}}}^{\prime }(4)=3\cdot 4^2\cdot \sqrt{4}+4^3\frac{1}{2\sqrt{4}}=112}$

• ${\displaystyle (\frac{7}{\exp }{)}^{\prime }(0)=7\cdot (\frac{1}{\exp }{)}^{\prime }(0)=7\cdot (-\frac{{\exp }^{\prime }(0)}{{\exp }^2(0)})=7\cdot (-\frac{\exp 0}{{\exp }^2(0)})=-7}$

• ${\displaystyle (\frac{\mathrm{X}}{\mathrm{X}-1}{)}^{\prime }(5)=\frac{{\mathrm{X}}^{\prime }(5)\cdot (\mathrm{X}-1)(5)-\mathrm{X}(5)\cdot (\mathrm{X}-1{)}^{\prime }(5)}{{(\mathrm{X}-1)}^2(5)}=\frac{1\cdot 4-5\cdot 1}{16}=-\frac{1}{16}}$

Proposition:  tan is differentiable at each ${\displaystyle a\ne (2k-1)\frac{\mathrm{\pi }}{2}}$ and cot at each ${\displaystyle a\ne k\mathrm{\pi }}$. Their derivative numbers calculate to

Proof:  The quotient rules guarantees the differentiability of tan and cot. We use Pythagoras' theorem (${\displaystyle {\sin }^2+{\cos }^2=1}$) to calculate

1. ►  ${\displaystyle {\tan }^{\prime }(a)=(\frac{\sin }{\cos }{)}^{\prime }(a)=\frac{{\sin }^{\prime }(a)\cdot \cos a-\sin a\cdot {\cos }^{\prime }(a)}{{\cos }^2(a)}=\frac{{\cos }^2(a)+{\sin }^2(a)}{{\cos }^2(a)}=\frac{1}{{\cos }^2(a)}}$

2. ►  ${\displaystyle {\cot }^{\prime }(a)=(\frac{\cos }{\sin }{)}^{\prime }(a)=\frac{{\cos }^{\prime }(a)\cdot \sin a-\cos a\cdot {\sin }^{\prime }(a)}{{\sin }^2(a)}=\frac{-{\sin }^2(a)-{\cos }^2(a)}{{\sin }^2(a)}=\frac{-1}{{\sin }^2(a)}}$

Proposition:  Let  ${\displaystyle f:B\to ℝ}$ and ${\displaystyle g:A\to ℝ}$ be any two functions, ${\displaystyle a\in A}$ an accumulation point of A such that ${\displaystyle g(a)}$ is a member and an accumulation point of B. If g is differentiable at a and  f differentiable at ${\displaystyle g(a)}$ then  ${\displaystyle f\circ g}$ is differentiable at a due to the

Proof:  We use the representation theorem [7.5.1] to find two functions ${\displaystyle r:A\to ℝ}$ and ${\displaystyle s:B\to ℝ}$, r continuous at a, s continuous at ${\displaystyle g(a)}$, such that ${\displaystyle r(a)=g^{\prime }(a),s(g(a))=f^{\prime }(g(a))}$ and

Thus the composit  ${\displaystyle f\circ g}$ could be calculated like this (note that ${\displaystyle \circ }$ is distributive from the right in respect to the first rules of arithmetic, X is the identity element and a constant function reproduces itself if it is the left input for ${\displaystyle \circ }$ ):

As r and s are continuous at the points mentioned above we see that ${\displaystyle t≔r\cdot s\circ (g(a)+(\mathrm{X}-a)\cdot r)}$ is continuous at a due to the calculation rules for continuous functions.  ${\displaystyle f\circ g}$ is thus differentiable at a according to the representation theorem with

Proposition:  For any ${\displaystyle n\in {\mathbb{N}}^{\ast }}$ we have: If  ${\displaystyle f:A\to ℝ}$ is differentiable at a, so is  ${\displaystyle f^n}$ with

Proof:  ${\displaystyle {\mathrm{X}}^n}$ is differentiable anywhere, especially at  ${\displaystyle f(a)}$, so that  ${\displaystyle f^n={\mathrm{X}}^n\circ f}$ is differentiable at a due to the chain rule. The derivative calculates as follows:

Example:  All functions involved are differentiable at the respective points. Thus their composits can be derivated using 8. and 9.

• ${\displaystyle ({\sin }^8{)}^{\prime }(0)=8\cdot {\sin }^7(0)\cdot {\sin }^{\prime }(0)=8\cdot 0\cdot 1=0}$

• ${\displaystyle (({\mathrm{X}}^3-3\mathrm{X}{)}^5{)}^{\prime }(2)=5(2^3-3\cdot 2{)}^4\cdot ({\mathrm{X}}^3-3\mathrm{X}{)}^{\prime }(2)=5\cdot 2^4\cdot (3\cdot 2^2-3)=720}$

• ${\displaystyle |{\mathrm{X}}^2-5{|}^{\prime }(1)=|\mathrm{X}{|}^{\prime }(-4)\cdot ({\mathrm{X}}^2-5{)}^{\prime }(1)=-2}$

• ${\displaystyle {\sqrt{|{\mathrm{X}}^2-5|}}^{\prime }}$