May 2018

We have defined the derivative of function $f(x)$ in “Inching towards Definite Integral” as

$f'(x) = \lim\limits_{h\to 0}{{f(x+h)-f(x)}\over{h}}$

An equivalent definition of $f'(x)$ is

$f'(x) = \lim\limits_{x^* \to x}{{f(x^*)-f(x)} \over {x^*-x}}$

We will prove the equivalency below:

$f'(x) = \lim\limits_{h \to 0}{{f(x+h)-f(x)} \over h}$

$\Longrightarrow \forall \epsilon >0, \exists \delta>0 \ni 0<|h-0|<\delta, |{{f(x+h)-f(x)} \over {h} }- f'(x)|< \epsilon\quad\quad\quad(1)$

Let $x^*=x+h$ , then $h = x^*-x$ , (1) becomes

$\forall \epsilon >0, \exists \delta>0 \ni 0<|(x^*-x)-0|<\delta, |{{f(x^*)-f(x)} \over {x^*-x}} - f'(x)|<\epsilon$

$\Longrightarrow \forall \epsilon >0, \exists \delta>0 \ni 0<|x^*-x|<\delta, |{{f(x^*)-f(x)} \over {x^*-x}}- f'(x)|<\epsilon$

$\Longrightarrow f'(x) = \lim\limits_{x^* \to x} {{f(x^*)-f(x)} \over {x^*-x}}$

Similarly,

$f'(x) = \lim\limits_{x^* \to x}{{f(x^*)-f(x)} \over {x^*-x}}$

$\Longrightarrow \forall \epsilon >0, \exists \delta>0 \ni 0<|x^*-x|<\delta, |{{f(x^*)-f(x)} \over {x^*-x} }- f'(x)|< \epsilon\quad\quad\quad(2)$

Let $h=x^*-x$ , then $x^*= x+h$ , (2) becomes

$\forall \epsilon >0, \exists \delta>0 \ni 0<|h|<\delta, |{{f(x+h)-f(x)} \over {h}} - f'(x)|<\epsilon$

$\Longrightarrow \forall \epsilon >0, \exists \delta>0 \ni 0<|h-0|<\delta, |{{f(x+h)-f(x)} \over {h}}- f'(x)|<\epsilon$

$\Longrightarrow f'(x) = \lim\limits_{h \to 0} {{f(x+h)-f(x)} \over {h}}$

Month: May 2018