Функције – извод функције 1

Текстови задатака објашњених у видео лекцији.

По дефиницији одредити изводе функција:

пр.1)   $f(x) = 3x - 1$
пр.2)   $f(x) = {x^3} - 5{x^2} + 4x - 1$
пр.3)   $f(x) = \frac{{2 - 3x}}{{x + 1}}$

пр.1)   $f(x) = 3x - 1$

$f'\left( x \right){\text{ }} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {x + \Delta x} \right){\text{  - }}f\left( x \right){\text{ }}}}{{\Delta x}}$

$f\left( {x + \Delta x} \right) = 3 \cdot \left( {x + \Delta x} \right) - 1$

$f'\left( x \right){\text{ }} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{3 \cdot \left( {x + \Delta x} \right) - 1 - \left( {3x{\text{ }} - {\text{ }}1} \right)}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{3x + 3\Delta x - 1 - 3x + 1}}{{\Delta x}} = $

$= \mathop {\lim }\limits_{\Delta x \to 0} \frac{{3\Delta x}}{{\Delta x}} = 3$

пр.2)   $f(x) = {x^3} - 5{x^2} + 4x - 1$

$f'\left( x \right){\text{ }} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {x + \Delta x} \right){\text{  - }}f\left( x \right){\text{ }}}}{{\Delta x}}$

$f\left( {x + \Delta x} \right) = {\left( {x + \Delta x} \right)^3} - 5{\left( {x + \Delta x} \right)^2} + 4\left( {x + \Delta x} \right) - 1 = $

$ = {x^3} + 3{x^2}\Delta x + 3x{\left( {\Delta x} \right)^2} + {\left( {\Delta x} \right)^3} - 5{x^2} - 10x\Delta x - 5\Delta x + 4{x^2} + 4\Delta x - 1$

$f'\left( x \right){\text{ }} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{x^3} + 3{x^2}\Delta x + 3x{{\left( {\Delta x} \right)}^2} + {{\left( {\Delta x} \right)}^3} - 5{x^2} - 10x\Delta x - 5\Delta x + 4{x^2} + 4\Delta x - 1 - \left( {{x^3} - 5{x^2} + 4x - 1} \right)}}{{\Delta x}} = $

$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta x\left( {3{x^2} + 3x\Delta x + \Delta x - 10x - 5\Delta x + 4} \right)}}{{\Delta x}} = 3{x^2} - 10x + 4$

пр.3)   $f(x) = \frac{{2 - 3x}}{{x + 1}}$

$f'\left( x \right){\text{ }} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {x + \Delta x} \right){\text{  - }}f\left( x \right){\text{ }}}}{{\Delta x}}$

$f\left( {x + \Delta x} \right) = \frac{{2 - 3\left( {x + \Delta x} \right)}}{{\left( {x + \Delta x} \right) + 1}}$

$f'\left( x \right){\text{ }} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {x + \Delta x} \right){\text{  - }}f\left( x \right){\text{ }}}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\frac{{2 - 3\left( {x + \Delta x} \right)}}{{x + \Delta x + 1}} - \frac{{2 - 3x}}{{x + 1}}}}{{\Delta x}} = $

$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\frac{{\left( {2 - 3\left( {x + \Delta x} \right)} \right)\left( {x + 1} \right) - \left( {2 - 3x} \right)\left( {x + \Delta x + 1} \right)}}{{\left( {x + \Delta x + 1} \right)\left( {x + 1} \right)}}}}{{\frac{{\Delta x}}{1}}} = $

$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left( {2 - 3\left( {x + \Delta x} \right)} \right)\left( {x + 1} \right) - \left( {2 - 3x} \right)\left( {x + \Delta x + 1} \right)}}{{\left( {x + \Delta x + 1} \right)\left( {x + 1} \right)\Delta x}} = $

$  = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{2x - 3{x^2} - 3x\Delta x + 2 - 3x - 3\Delta x - 2x - 2\Delta x - 2 + 3{x^2} + 3x\Delta x + 3x}}{{\left( {x + \Delta x + 1} \right)\left( {x + 1} \right)\Delta x}} =  $

$ = \frac{{ - 5}}{{{{\left( {x + 1} \right)}^2}}}$