Рационал функциялардың туындысы
2.1. Берілген функциялардың туындыларын есептеңіз.
1) $f(x)=x^8-3 x^4-x+5$.
Шешуі
$${\left( {{x^p}} \right)^\prime } = p \cdot {x^{p - 1}}$$ $$f'(x) = {\left( {{x^8}} \right)^\prime } - 3{\left( {{x^4}} \right)^\prime } - (x)' + (5)' = $$ $$ = 8{x^7} - 3 \cdot 4{x^3} - 1 + 0 = 8{x^7} - 12{x^3} - 1$$
2) $f(x)=\dfrac{x^5}{5}-\dfrac{2}{3} x^3-3(x+1)$.
Шешуі
$$f'(x) = \frac{1}{5} \cdot {\left( {{x^5}} \right)^\prime } - \frac{2}{3} \cdot {\left( {{x^3}} \right)^\prime } - 3 \cdot (x + 1)' = $$ $$ = \frac{1}{5} \cdot 5{x^4} - \frac{2}{3} \cdot 3{x^2} - 3 \cdot 1 = {x^4} - 2{x^2} - 3$$
3) $f(x)=\dfrac{1}{2} x^2-x \sqrt{3}+2^{\sqrt{2}}$ .
Шешуі
$$f^{\prime}(x)=\frac{1}{2} \cdot\left(x^2\right)^{\prime}-\sqrt{3} \cdot(x)^{\prime}+\left(2^{\sqrt{2}}\right)^{\prime}=\frac{1}{2} \cdot 2 x-\sqrt{3} \cdot 1+0=x-\sqrt{3}$$
4) $f(x)=\dfrac{4}{x}-\dfrac{4}{x^2}$ .
Шешуі
$$f(x)=\frac{4}{x}-\frac{4}{x^2}=\frac{4}{x}-4 \cdot x^{-2}$$ $$f^{\prime}(x)=\left(\frac{4}{x}\right)^{\prime}-4 \cdot\left(x^{-2}\right)^{\prime}=-\frac{4}{x^2}-4 \cdot(-2) \cdot x^{-3}=-\frac{4}{x^2}+\frac{8}{x^3}$$
5) $f(x)=\left(x^4+\dfrac{1}{x^3}\right) \cdot x^4$.
Шешуі
$$f(x)=x^8+x$$ $$f^{\prime}(x)=8 x^7+1$$
6) $f(x)=\dfrac{x^2+x-3}{2 x}$.
Шешуі
$$ \boxed{\left(\frac{f(x)}{g(x)}\right)^{\prime}=\frac{f^{\prime}(x) \cdot g(x)-f(x) \cdot g^{\prime}(x)}{g^2(x)}} $$ $$ f^{\prime}(x)=\frac{\left(x^2+x-3\right)^{\prime} \cdot 2 x-\left(x^2+x-3\right) \cdot(2 x)^{\prime}}{(2 x)^2} $$ $$ f^{\prime}(x)=\frac{1}{2}(x)^{\prime}+\left(\frac{1}{2}\right)^{\prime}-\frac{3}{2}\left(\frac{1}{x}\right)^{\prime}=\frac{1}{2}+\frac{3}{2 x^2}=\frac{x^2+3}{2 x^2} $$
7) $f(x)=2|x|+1$.
Шешуі
$$f(x) = 2\left| x \right| + 1 = \left\{ {\begin{array}{*{20}{c}}{2x + 1,}&{x \ge 0}\\{ - 2x + 1,}&{x \lt 0}\end{array}} \right.$$ $f(x)=2|x|+1$ функциясының $x=0$ нүктесінде туындысы жоқ. Сондықтан, $$f'\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{{{\left( {2x + 1} \right)}^\prime }}&{x \gt 0}\\{{{\left( { - 2x + 1} \right)}^\prime }}&{x \lt 0}\end{array} = \left\{ {\begin{array}{*{20}{c}}{2,}&{x \gt 0}\\{ - 2}&{x \lt 0}\end{array}} \right.} \right.$$
8) $f(x)=x|x|$.
Шешуі
$$ f(x)=x|x|=\left\{\begin{aligned} x^2,& \quad x \geq 0 \\ -x^2,& \quad x \lt 0 \end{aligned}\right. $$ $$ f^{\prime}(x)=\left\{\begin{array}{cc} \left(x^2\right)^{\prime},& x \geq 0 \\ \left(-x^2\right)^{\prime},& x \lt 0 \end{array}=\left\{\begin{array}{cc} 2 x,& x \geq 0 \\ -2 x,& x \lt 0 \end{array}\right.\right. $$ Бұл функцияның $x=0$ нүктесінде туындысы бар және ол 0-ге тең.
2.2 $\quad x_0$ нүктесіндегі берілген функциялардың туындысын есептеңіз. .
1) $f(x)=\left(x^2-1\right)(2-3 x), \quad f^{\prime}(2)-?$
Шешуі
$$ f^{\prime}(x)=\left(2 x^2-2-3 x^3+3 x\right)^{\prime}=4 x-9 x^2+3 $$ $$ f^{\prime}(2)=4 \cdot 2-9 \cdot 2^2+3=-25 $$
2) $f(x)=\dfrac{2-3 x}{x-1},\quad x=2$.
Шешуі
$$ f^{\prime}(x)=\left(\frac{2-3 x}{x-1}\right)^{\prime}=\frac{(2-3 x)^{\prime} \cdot(x-1)-(x-1)^{\prime} \cdot(2-3 x)}{(x-1)^2}= $$ $$ =\frac{-3 \cdot(x-1)-1 \cdot(2-3 x)}{(x-1)^2}=\frac{-3 x+3-2+3 x}{(x-1)^2}=\frac{1}{(x-1)^2} $$ $$ x=2 \text{ нүктесінде }: \quad f^{\prime}(2)=\frac{1}{(2-1)^2}=1 $$
3) $f(x)=\dfrac{1}{|x|}, \quad f^{\prime}(-2)-?$
Шешуі
$$ f(x)=\frac{1}{|x|}=\left\{\begin{array}{ll} \frac{1}{x},& x\gt 0 \\ -\frac{1}{x},& x\lt 0 \end{array}\right. $$ $$ f^{\prime}(x)=\left\{\begin{array}{l} \left(\frac{1}{x}\right)^{\prime}, \quad x\gt 0, \\ \left(-\frac{1}{x}\right)^{\prime}, \quad x\lt 0 ; \end{array}=\left\{\begin{array}{ll} \frac{1}{x^2},& \quad x\gt 0 \\ \frac{1}{x^2},& \quad x\lt 0 \end{array}\right.\right. $$ $$ x=-2\lt 0: \quad f^{\prime}(-2)=\frac{1}{(-2)^2}=\frac{1}{4} $$
2.3 Күрделі функциялардың туындысын есептеңіз. .
1) $f(x)=\left(x^7-3 x^4\right)^{120}$.
Шешуі
$$ f(u)=u^{120}, \quad u(x)=x^7-3 x^4 $$ $$ f^{\prime}(x)=\left(\left(x^7-3 x^4\right)^{120}\right)^{\prime}=120\left(x^7-3 x^4\right)^{119} \cdot\left(x^7-3 x^4\right)^{\prime}= $$ $$ =120\left(x^7-3 x^4\right)^{119}\left(7 x^6-12 x^3\right) $$
2) $f(x)=\left(2 x \cdot \sin \dfrac{\pi}{4}+1\right)^2, \quad f^{\prime}(\sqrt{2})-?$
Шешуі
$$ f(x)=\left(2 x \cdot \sin \frac{\pi}{4}+1\right)^2=\left(2 x \cdot \frac{\sqrt{2}}{2}+1\right)^2=(x \sqrt{2}+1)^2 $$ $$ f^{\prime}(x)=\left((x \sqrt{2}+1)^2\right)^{\prime}=2(x \sqrt{2}+1) \cdot(x \sqrt{2}+1)^{\prime}=2 \sqrt{2}(x \sqrt{2}+1) $$
3) $f(x)=\dfrac{3}{x^2+x+1}, \quad f^{\prime}(1)-?$
Шешуі
$$ f^{\prime}(\sqrt{2})=6 \sqrt{2} $$ $$ f^{\prime}(x)=\left(\frac{3}{x^2+x+1}\right)^{\prime}=3 \cdot \frac{-1}{\left(x^2+x+1\right)^2} \cdot\left(x^2+x+1\right)^{\prime}=-\frac{3(2 x+1)}{\left(x^2+x+1\right)^2}= $$ $$ =-\frac{6 x+3}{\left(x^2+x+1\right)^2} $$ $$ f^{\prime}(1)=-\frac{6 \cdot 1+3}{3^2}=-1 $$
4) $f(x)=\dfrac{1}{\left(x^2-1\right)^4}$.
Шешуі
$$ f(x)=\frac{1}{\left(x^2-1\right)^4}=\left(x^2-1\right)^{-4} $$ $$ f^{\prime}(x)=\left(\left(x^2-1\right)^{-4}\right)^{\prime}=-4\left(x^2-1\right)^{-5} \cdot\left(x^2-1\right)^{\prime}=-4 \cdot \frac{1}{\left(x^2-1\right)^5} \cdot 2 x=\frac{-8 x}{\left(x^2-1\right)^5} $$