Функције – извод функције 2

Текстови задатака објашњених у видео лекцији.

По дефиницији одредити изводе функција:

пр.4)   $f(x) = \sqrt {4x - 1} $

пр.5)   $f(x) = {e^x}$

пр.6)   $f(x) = \ln x$

пр.4)   $f(x) = \sqrt {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) = \sqrt {4\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{{\sqrt {4\left( {x + \Delta x} \right) - 1}  - \sqrt {4x - 1} }}{{\Delta x}} \cdot \frac{{\sqrt {4\left( {x + \Delta x} \right) - 1}  + \sqrt {4x - 1} }}{{\sqrt {4\left( {x + \Delta x} \right) - 1}  + \sqrt {4x - 1} }} = $

$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{{{\sqrt {4\left( {x + \Delta x} \right) - 1} }^2} - {{\sqrt {4x - 1} }^2}}}{{\Delta x\left( {\sqrt {4\left( {x + \Delta x} \right) - 1}  + \sqrt {4x - 1} } \right)}} = $

$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{4x + 4\Delta x - 1 - 4x + 1}}{{\Delta x\left( {\sqrt {4\left( {x + \Delta x} \right) - 1}  + \sqrt {4x - 1} } \right)}} = \frac{4}{{\sqrt {4x - 1}  + \sqrt {4x - 1} }} = \frac{2}{{\sqrt {4x - 1} }}$

пр.5)   $f(x) = {e^x}$

$f\left( {x + \Delta x} \right) = {e^{x + \Delta x}} = {e^x} \cdot {e^{\Delta x}}$

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

пр.6)   $f(x) = \ln x$

$f\left( {x + \Delta x} \right) = \ln \left( {x + \Delta x} \right)$

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

$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {1 + \frac{{\Delta x}}{x}} \right)}}{{\frac{{\Delta x}}{x} \cdot x}} = \frac{1}{x}$