Иррационал функциялардың туындысы

Иррационал функциялардың сы

2.4. Келесі функциялардың туындысын табыңыз.

1) $f(x)=2,5 x^2+20 \sqrt{x}-3 \cdot \sqrt[3]{x}$.

Шешуі

$f(x)=2,5 x^2+20 \sqrt{x}-3 \cdot \sqrt[3]{x}$
$\sqrt[3]{x}=x^{\frac{1}{3}} $, онда
$f^{\prime}(x)=2,5\left(x^2\right)^{\prime}+20(\sqrt{x})^{\prime}-3\left(x^{\frac{1}{3}}\right)^{\prime}=2,5 \cdot 2 x+20 \cdot \dfrac{1}{2 \sqrt{x}}-3 \cdot \dfrac{1}{3} x^{-\frac{2}{3}}=$
$=5 x+\dfrac{10}{\sqrt{x}}-\dfrac{1}{\sqrt[3]{x^2}}$


2) $f(x)=5 x \cdot \sqrt[5]{x^4}-(\sqrt{\pi})^3$.

Шешуі

$f(x)=5 x \cdot \sqrt[5]{x^4}-(\sqrt{\pi})^3=5 x^1 \cdot x^{\frac{4}{5}}-(\sqrt{\pi})^3=5 x^{\frac{9}{5}}-(\sqrt{\pi})^3$
$(\sqrt{\pi})^3$– const екенін ескерсек,
$f^{\prime}(x)=5\left(x^{\frac{9}{5}}\right)^{\prime}-\left((\sqrt{\pi})^3\right)^{\prime}=5 \cdot \frac{9}{5} \cdot x^{\frac{4}{5}}-0=9 \cdot \sqrt[5]{x^4}$.


3) $f(x)=\dfrac{x^2+2 x}{\sqrt{x}}$.

Шешуі

$f(x)=\dfrac{x^2+2 x}{\sqrt{x}}=\dfrac{x^2}{\sqrt{x}}+\dfrac{2 x}{\sqrt{x}}=x \sqrt{x}+2 \sqrt{x}=x^{\frac{3}{2}}+2 x^{\frac{1}{2}}$
$f^{\prime}(x)=\left(x^{\frac{3}{2}}\right)^{\prime}+2\left(x^{\frac{1}{2}}\right)^{\prime}=\dfrac{3}{2} \cdot x^{\frac{1}{2}}+2 \cdot \dfrac{1}{2} x^{-\frac{1}{2}}=\dfrac{3}{2} \sqrt{x}+\dfrac{1}{\sqrt{x}}$


4) $f(x)=\frac{1}{4} \cdot \sqrt[4]{8 x^3 \cdot \sqrt{x}}$.

Шешуі

$f(x)=\dfrac{1}{4} \cdot \sqrt[4]{8 x^3 \cdot \sqrt{x}}=\dfrac{1}{4} \cdot \sqrt[4]{8} \cdot\left(x^3 \cdot x^{\frac{1}{2}}\right)^{\frac{1}{4}}=\dfrac{\sqrt[4]{8}}{4} \cdot\left(x^{\frac{7}{2}}\right)^{\frac{1}{4}}=\dfrac{\sqrt[4]{8}}{4} \cdot x^{\frac{7}{8}}$
$f^{\prime}(x)=\left(\dfrac{\sqrt[4]{8}}{4} \cdot x^{\frac{7}{8}}\right)^{\prime}=\dfrac{\sqrt[4]{8}}{4} \cdot \dfrac{7}{8} \cdot x^{-\frac{1}{8}}=\dfrac{7 \cdot \sqrt[4]{8}}{32 \cdot \sqrt[8]{x}}$


5) $f(x)=\dfrac{1}{\sqrt[3]{x^2}}+\dfrac{1}{\sqrt[4]{x^3}}$.

Шешуі

$f(x)=\dfrac{1}{\sqrt[3]{x^2}}+\dfrac{1}{\sqrt[4]{x^3}}=x^{-\frac{2}{3}}+x^{-\frac{3}{4}}$
$f^{\prime}(x)=\left(x^{-\frac{2}{3}}\right)^{\prime}+\left(x^{-\frac{3}{4}}\right)^{\prime}=-\dfrac{2}{3} x^{-\frac{5}{3}}-\dfrac{3}{4} x^{-\frac{7}{4}}=-\dfrac{2}{3 \cdot \sqrt[3]{x^5}}-\dfrac{3}{4 \cdot \sqrt[4]{x^7}}=$
$=-\dfrac{2}{3 x \cdot \sqrt[3]{x^2}}-\dfrac{3}{4 x \cdot \sqrt[4]{x^3}}=-\dfrac{2 \cdot \sqrt[3]{x}}{3 x^2}-\dfrac{3 \cdot \sqrt[4]{x}}{4 x^2}=-\dfrac{8 \cdot \sqrt[3]{x}+9 \cdot \sqrt[4]{x}}{12 x^2}$


2.5 Күрделі функциялардың туындысын есептеңіз.

1) $f(x)=x^3 \cdot \sqrt{x-1}, \quad f^{\prime}(2)-?$

Шешуі

$f(x)=x^3 \cdot \sqrt{x-1}=\sqrt{x^6(x-1)}=\sqrt{x^7-x^6}$
$f^{\prime}(x)=\left(\sqrt{x^7-x^6}\right)^{\prime}=\frac{1}{2 \sqrt{x^7-x^6}} \cdot\left(x^7-x^6\right)^{\prime}=$
$=\dfrac{7 x^6-6 x^5}{2 \sqrt{x^7-x^6}}=\dfrac{x^5(7 x-6)}{2 x^3 \sqrt{x-1}}=\dfrac{7 x^3-6 x^2}{2 \sqrt{x-1}}$
$f^{\prime}(2)=16$


2) $f(x)=\sqrt{5-x^2}+\dfrac{1}{(3-x)^2}, \quad f^{\prime}(2)-?$

Шешуі

$f(x)=\sqrt{5-x^2}+\dfrac{1}{(3-x)^2}=\sqrt{5-x^2}+(3-x)^{-2}$
$f^{\prime}(x)=\left(\sqrt{5-x^2}\right)^{\prime}+\left((3-x)^{-2}\right)^{\prime}=\dfrac{1}{2 \sqrt{5-x^2}} \cdot\left(5-x^2\right)^{\prime}+$
$+(-2)(3-x)^{-3} \cdot(3-x)^{\prime}=\dfrac{-2 x}{2 \sqrt{5-x^2}}-\dfrac{2(-1)}{(3-x)^3}=-\dfrac{x}{\sqrt{5-x^2}}+\dfrac{2}{(3-x)^3}$
$f^{\prime}(2)=-2+2=0$.


3) $f(x)=\sqrt[3]{\left(x^3+1\right)^2}$ .

Шешуі

$f(x)=\sqrt[3]{\left(x^3+1\right)^2}=\left(x^3+1\right)^{\frac{2}{3}}$
$f^{\prime}(x)=\left(\left(x^3+1\right)^{\frac{2}{3}}\right)^{\prime}=\dfrac{2}{3}\left(x^3+1\right)^{-\frac{1}{3}} \cdot\left(x^3+1\right)^{\prime}=\dfrac{2}{3} \cdot \dfrac{1}{\sqrt[3]{x^3+1}} \cdot 3 x^2=\dfrac{2 x^2}{\sqrt[3]{x^3+1}}$


4) $f(x)=\dfrac{2 x-1}{\sqrt{x^2+1}}$ .

Шешуі

$f(x)=\dfrac{2 x-1}{\sqrt{x^2+1}}$
$f^{\prime}(x)=\dfrac{(2 x-1)^{\prime} \cdot \sqrt{x^2+1}-\left(\sqrt{x^2+1}\right)^{\prime}(2 x-1)}{\left(\sqrt{x^2+1}\right)^2}=\dfrac{2 \sqrt{x^2+1}-\frac{1}{2 \sqrt{x^2+1}}\left(x^2+1\right)^{\prime} \cdot(2 x-1)}{x^2+1}=$
$=\dfrac{2 \sqrt{x^2+1}-\frac{2 x^2-x}{\sqrt{x^2+1}}}{x^2+1}=\dfrac{\frac{2\left(x^2+1\right)-2 x^2+x}{\sqrt{x^2+1}}}{x^2+1}=\dfrac{x+2}{\left(x^2+1\right) \sqrt{x^2+1}}=$
$=\dfrac{x+2}{\sqrt{\left(x^2+1\right)^2\left(x^2+1\right)}}=\dfrac{x+2}{\sqrt{\left(x^2+1\right)^3}}$




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Сіздің электронды почтаңыз жарияланбайды, Міндетті жолдарды толтырып шығыңыз.