Infi-1 21

Differentiability #definition

\begin{matrix} f^{'} (a) = lim_{x \to a} \frac{f (x) - f (a)}{x - a} \\ f^{'} (a) = lim_{h \to 0} \frac{f (a + h) - f (a)}{h} \end{matrix}

\begin{matrix} lim_{x \to a} f (x) = L ⟺ \forall x_{n} : x_{n} \to a \land x_{n} \neq a ⟹ f (x_{n}) \to L \end{matrix}

\begin{matrix} If function is differentiable at a, then it is also continuous at a \end{matrix}

\begin{matrix} (\sin (x^{2}))^{'} = \cos (x^{2}) \cdot 2 x \\ (f \circ g)^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) \end{matrix}

\begin{matrix} Let f, g functions \\ g is differentiable at a \\ f is differentiable at g (a) \\ ⟹ (f \circ g) is differentiable at a \\ and (f \circ g)^{'} (a) = f^{'} (g (a)) \cdot g^{'} (a) \\ Proof: \\ (f \circ g)^{'} (a) = lim_{x \to a} \frac{(f \circ g) (x) - (f \circ g) (a)}{x - a} = lim_{x \to a} \frac{f (g (x)) - f (g (a))}{x - a} = \\ = lim_{x \to a} \frac{(f (g (x)) - f (g (a))) (g (x) - g (a))}{(g (x) - g (a)) (x - a)} = lim_{x \to a} \frac{f (g (x)) - f (g (a))}{g (x) - g (a)} \cdot \frac{g (x) - g (a)}{x - a} = \\ = \underset{x \to a, g is continuous at a ⟹ g (x) \to g (a)}{\underset{⏟}{lim_{x \to a} \frac{f (g (x)) - f (g (a))}{g (x) - g (a)}}} \cdot lim_{x \to a} \frac{g (x) - g (a)}{x - a} = f^{'} (g (a)) \cdot g^{'} (a) \end{matrix}

\begin{matrix} \log_{a} (x) = \frac{\ln x}{\ln a} = \ln x \cdot \frac{1}{\ln a} \\ (\ln x)^{'} = \frac{1}{x} \\ ⟹ (\log_{a} (x))^{'} = \frac{1}{x \ln a} \end{matrix}

\begin{matrix} (\ln x)^{'} = \frac{1}{x} \\ Proof: \\ \frac{1}{t} \to \infty ⟹ (1 + t)^{1 / t} = {(1 + \frac{1}{\frac{1}{t}})}^{1 / t} \to e \\ lim_{t \to 0} \frac{1}{t} \cdot \ln (1 + t) = lim_{t \to 0} \ln ((1 + t)^{1 / t}) = lim_{t \to 0} \ln (e) = 1 \\ ⟹ lim_{t \to 0} \frac{\ln (1 + t)}{t} = 1 \\ (\ln x)^{'} = lim_{h \to 0} \frac{\ln (x + h) - \ln (x)}{h} = lim_{h \to 0} \frac{\ln (\frac{x + h}{x})}{h} = lim_{h \to 0} \frac{\ln (1 + \frac{h}{x})}{h} = \\ = lim_{h \to 0} \frac{\ln (1 + \frac{h}{x})}{\frac{h}{x}} \cdot \frac{1}{x} = 1 \cdot \frac{1}{x} = \frac{1}{x} \end{matrix}

\begin{matrix} (a^{x})^{'} = a^{x} \ln a \\ (e^{x})^{'} = e^{x} \\ Proof: \\ t \to 0 ⟹ a^{t} \to 1 ⟹ w \to 0 \\ w = a^{t} - 1 \\ a^{t} = w + 1 ⟹ \ln (a^{t}) = \ln (w + 1) ⟹ t = \frac{\ln (w + 1)}{\ln (a)} \\ lim_{t \to 0} \frac{a^{t} - 1}{t} = lim_{w \to 0} \frac{w \ln (a)}{\ln (w + 1)} = \ln (a) \cdot lim_{w \to 0} \frac{w}{\ln (w + 1)} = \ln (a) \cdot lim_{w \to 0} \frac{1}{\frac{\ln (w + 1)}{w}} = \ln (a) \\ ⟹ lim_{t \to 0} \frac{a^{t} - 1}{t} = \ln a \\ (a^{x})^{'} = lim_{h \to 0} \frac{a^{x + h} - a^{x}}{h} = lim_{h \to 0} \frac{a^{x} (a^{h} - 1)}{h} = a^{x} lim_{h \to 0} \frac{a^{h} - 1}{h} = a^{x} \ln a \\ (e^{x})^{'} = e^{x} \ln e = e^{x} \end{matrix}

\begin{matrix} (x^{a})^{'} = a x^{a - 1} \\ Explanation: \\ (x^{a})^{'} = (e^{\ln (x^{a})})^{'} = e^{\ln (x^{a})} \cdot \underset{(a \ln x)^{'} = a (\ln x)^{'}}{\underset{⏟}{(\ln (x^{a}))^{'}}} = e^{\ln (a^{x})} \cdot \frac{a}{x} = x^{a} \cdot a \cdot \frac{1}{x} = a \cdot x^{a - 1} \end{matrix}

\begin{matrix} {(\frac{f}{g})}^{'} = \frac{f^{'} g - f g^{'}}{g^{2}} \\ (\tan x)^{'} = {(\frac{\sin x}{\cos x})}^{'} = \frac{\cos^{2} (x) + \sin^{2} (x)}{\cos^{2} (x)} = \frac{1}{\cos^{2} (x)} \\ Let f, g functions \\ f, g differentiable at a \\ ⟹ \frac{f}{g} is differentiable at a \\ and {(\frac{f}{g})}^{'} (a) = \frac{f^{'} (a) g (a) - f (a) g^{'} (a)}{(g (a))^{2}} \\ Proof: \\ {(\frac{f}{g})}^{'} = (f \cdot g^{- 1})^{'} = f^{'} g^{- 1} + f (g^{- 1})^{'} = f^{'} \cdot g^{- 1} + f \cdot (- 1) \cdot g^{- 1 - 1} \cdot g^{'} = \\ = \frac{f^{'} g}{g^{2}} - \frac{f g^{'}}{g^{2}} = \frac{f^{'} g - f g^{'}}{g^{2}} \end{matrix}

\begin{matrix} f^{g} = e^{\ln (f^{g})} = e^{g \ln f} \end{matrix}