Infi-1 20

Derivative #definition

\begin{matrix} Let a \in D o m (f) \\ Function is called differentiable at a if exists \\ f^{'} (a) = lim_{h \to 0} \frac{f (a + h) - f (a)}{h} \end{matrix}

\begin{matrix} Definition can be written as: \\ f^{'} (x) = lim_{x \to a} \frac{f (x) - f (a)}{x - a} \end{matrix}

\begin{matrix} If f is differentiable at a, then it is continuous at a \\ Proof: \\ \exists f^{'} (a) ⟹ \exists lim_{x \to a} \frac{f (x) - f (a)}{x - a} \\ x - a \to 0 ⟹ f (x) - f (a) \to 0 ⟹ f (x) \to f (a) \\ ⟹ f is continuous at a \end{matrix}

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

\begin{matrix} x \in R \\ (\sin x)^{'} = \cos x \\ (\cos x)^{'} = - \sin x \\ Proof: \\ lim_{t \to 0} \frac{\sin t}{t} = 1 \\ lim_{t \to 0} \frac{1 - \cos t}{t} = 0 \\ \sin (x + h) = \sin x \cos h + \sin h \cos x \\ \cos (x + h) = \cos x \cosh - \sin x \sin h \\ (\sin x)^{'} = lim_{h \to 0} \frac{\sin (x + h) - \sin x}{h} = lim_{h \to 0} \frac{\sin x \cos h + \sin h \cos x - \sin x}{h} = \\ = lim_{h \to 0} \frac{\sin x (\cos h - 1) + \sin h \cos x}{h} = lim_{h \to 0} \sin x \underset{\to 0}{\underset{⏟}{\frac{(\cos h - 1)}{h}}} + \cos x \underset{\to 1}{\underset{⏟}{\frac{(\sin h)}{h}}} = \cos x \\ (\cos x)^{'} = lim_{h \to 0} \frac{\cos (x + h) - \cos x}{h} = lim_{h \to 0} \frac{\cos x \cos h - \sin h \sin x - \cos x}{h} = \\ = lim_{h \to 0} \frac{\cos x (\cos h - 1) + \sin h \sin x}{h} = lim_{h \to 0} \cos x \underset{\to 0}{\underset{⏟}{\frac{(\cos h - 1)}{h}}} - \sin x \underset{\to 1}{\underset{⏟}{\frac{(\sin h)}{h}}} = - \sin x \end{matrix}

\begin{matrix} Let f, g functions \\ a, C \in R \\ If f, g are differentiable at a, then f + g is also differentioable at a and: \\ (f + g)^{'} (a) = f^{'} (a) + g^{'} (a) \\ If f is differentiable at a, then C f is also differentiable at a and: \\ (C f)^{'} (a) = C f^{'} (a) \\ Proof: \\ (f + g)^{'} (a) = lim_{x \to a} \frac{(f + g) (x) - (f + g) (a)}{x - a} = lim_{x \to a} \frac{f (x) + g (x) - (f (a) + g (a))}{x - a} = \\ = lim_{x \to a} \frac{f (x) - f (a)}{x - a} + lim_{x \to a} \frac{g (x) - g (a)}{x - a} = f^{'} (a) + g^{'} (a) \\ (C f)^{'} (a) = lim_{x \to a} \frac{(C f) (x) - (C f) (a)}{x - a} = lim_{x \to a} \frac{C f (x) - C f (a)}{x - a} = C lim_{x \to a} \frac{f (x) - f (a)}{x - a} = \\ = C f^{'} (a) \end{matrix}

\begin{matrix} Let f, g functions \\ a \in R \\ If f, g are differentiable at a, then f \cdot g is also differentiable at a and: \\ (f \cdot g)^{'} (a) = f^{'} (a) g (a) + f (a) g^{'} (a) \\ Proof: \\ (f \cdot g)^{'} (a) = lim_{h \to 0} \frac{(f \cdot g) (a + h) - (f \cdot g) (a)}{h} = lim_{h \to 0} \frac{f (a + h) \cdot g (a + h) - f (a) \cdot g (a)}{h} = \\ = lim_{h \to 0} \frac{f (a + h) \cdot g (a + h) - f (a) g (a + h) + f (a) g (a + h) - f (a) \cdot g (a)}{h} = \\ = lim_{h \to 0} \frac{g (a + h) (f (a + h) - f (a)) + f (a) (g (a + h) - g (a))}{h} = \\ = lim_{h \to 0} \frac{g (a + h) (f (a + h) - f (a)) + f (a) (g (a + h) - g (a))}{h} = \\ = lim_{h \to 0} (\underset{\to g (a)}{\underset{⏟}{g (a + h)}} \frac{f (a + h) - f (a)}{h} + f (a) \frac{g (a + h) - g (a)}{h}) = \\ = f^{'} (a) g (a) + f (a) g^{'} (a) \end{matrix}

\begin{matrix} h \to 0 ⟹ a + h \to a \\ g is differentiable at a ⟹ g is continuous at a \\ ⟹ [a + h \to a ⟹ g (a + h) \to g (a)] \end{matrix}

\begin{matrix} (x^{3} \cos x)^{'} = (x^{3})^{'} \cos x + x^{3} (\cos x)^{'} = 3 x^{2} \cos x - x^{3} \sin x \end{matrix}