вправа 19.12 гдз 10 клас математика Мерзляк Номіровський 2018
Вправа 19.12
Умова:
Умова:
Обчисліть значення похідної функції f у точці х0.
Відповідь ГДЗ:
\begin{equation} 1){f}'(x)={\left ( x\sqrt{x} \right )}'= \end{equation} \begin{equation} =\left ( \sqrt{x^{3}} \right )={\left ( x\frac{3}{2} \right )}'= \end{equation} \begin{equation} =\frac{3}{2}x^{\frac{3}{2}-1}=\frac{3}{2}x^{\frac{1}{2}}= \end{equation} \begin{equation} =\frac{3\sqrt{x}}{2}; \end{equation} \begin{equation} {f}'(81)=\frac{3\sqrt{81}}{2}= \end{equation} \begin{equation} =\frac{3 \cdot 9}{2}=\frac{27}{2}=13,5; \end{equation} \begin{equation} 2){f}'(x)={\left ( x^{3}\sqrt[4]{x} \right )}'= \end{equation} \begin{equation} ={\left ( \sqrt[4]{x^{13}} \right )}'=\frac{13}{4}x^{\frac{13}{4}-1}= \end{equation} \begin{equation} =\frac{13}{4}x^{\frac{9}{4}}=\frac{13\sqrt[4]{x^{9}}}{4}; \end{equation} \begin{equation} {f}'(x)=\frac{13 \cdot 1}{4}=\frac{13}{4}; \end{equation} \begin{equation} 3){f}'(x)={\left ( \sqrt{x\sqrt{x}} \right )}'= \end{equation} \begin{equation} ={\left ( \sqrt{\sqrt{3}} \right )}'={\left ( \sqrt[4]{x^{3}} \right )}'= \end{equation} \begin{equation} ={\left ( x^{\frac{3}{4}} \right )}'=\frac{3}{4}x^{\frac{3}{4}-1}= \end{equation} \begin{equation} =\frac{3}{4}x^{-\frac{1}{4}}=\frac{3}{4\sqrt[4]{x}}; \end{equation} \begin{equation} {f}'(16)=\frac{3}{4\sqrt[4]{16}}= \end{equation} \begin{equation} =\frac{3}{4 \cdot 2}=\frac{3}{8}; \end{equation} \begin{equation} 4){f}'(x)={\left ( \frac{x^{2}}{\sqrt[6]{x}} \right )}'= \end{equation} \begin{equation} ={\left ( x^{2-\frac{1}{6}} \right )}'={\left ( x^{\frac{11}{6}} \right )}'= \end{equation} \begin{equation} =\frac{11}{6} \cdot x^{\frac{11}{6}-1}= \end{equation} \begin{equation} =\frac{11}{6} \cdot x^{\frac{6}{6}}= \end{equation} \begin{equation} =\frac{11 \cdot \sqrt[6]{x^{5}}}{6}; \end{equation} \begin{equation} {f}'(64)=\frac{11 \cdot \sqrt[6]{64^{5}}}{6}= \end{equation} \begin{equation} =\frac{11 \cdot \sqrt[6]{2^{30}}}{6}= \end{equation} \begin{equation} =\frac{11 \cdot 2^{5}}{6}=\frac{11 \cdot 32}{6}= \end{equation} \begin{equation} =\frac{352}{6}=\frac{176}{3}. \end{equation}
\begin{equation} 1){f}'(x)={\left ( x\sqrt{x} \right )}'= \end{equation} \begin{equation} =\left ( \sqrt{x^{3}} \right )={\left ( x\frac{3}{2} \right )}'= \end{equation} \begin{equation} =\frac{3}{2}x^{\frac{3}{2}-1}=\frac{3}{2}x^{\frac{1}{2}}= \end{equation} \begin{equation} =\frac{3\sqrt{x}}{2}; \end{equation} \begin{equation} {f}'(81)=\frac{3\sqrt{81}}{2}= \end{equation} \begin{equation} =\frac{3 \cdot 9}{2}=\frac{27}{2}=13,5; \end{equation} \begin{equation} 2){f}'(x)={\left ( x^{3}\sqrt[4]{x} \right )}'= \end{equation} \begin{equation} ={\left ( \sqrt[4]{x^{13}} \right )}'=\frac{13}{4}x^{\frac{13}{4}-1}= \end{equation} \begin{equation} =\frac{13}{4}x^{\frac{9}{4}}=\frac{13\sqrt[4]{x^{9}}}{4}; \end{equation} \begin{equation} {f}'(x)=\frac{13 \cdot 1}{4}=\frac{13}{4}; \end{equation} \begin{equation} 3){f}'(x)={\left ( \sqrt{x\sqrt{x}} \right )}'= \end{equation} \begin{equation} ={\left ( \sqrt{\sqrt{3}} \right )}'={\left ( \sqrt[4]{x^{3}} \right )}'= \end{equation} \begin{equation} ={\left ( x^{\frac{3}{4}} \right )}'=\frac{3}{4}x^{\frac{3}{4}-1}= \end{equation} \begin{equation} =\frac{3}{4}x^{-\frac{1}{4}}=\frac{3}{4\sqrt[4]{x}}; \end{equation} \begin{equation} {f}'(16)=\frac{3}{4\sqrt[4]{16}}= \end{equation} \begin{equation} =\frac{3}{4 \cdot 2}=\frac{3}{8}; \end{equation} \begin{equation} 4){f}'(x)={\left ( \frac{x^{2}}{\sqrt[6]{x}} \right )}'= \end{equation} \begin{equation} ={\left ( x^{2-\frac{1}{6}} \right )}'={\left ( x^{\frac{11}{6}} \right )}'= \end{equation} \begin{equation} =\frac{11}{6} \cdot x^{\frac{11}{6}-1}= \end{equation} \begin{equation} =\frac{11}{6} \cdot x^{\frac{6}{6}}= \end{equation} \begin{equation} =\frac{11 \cdot \sqrt[6]{x^{5}}}{6}; \end{equation} \begin{equation} {f}'(64)=\frac{11 \cdot \sqrt[6]{64^{5}}}{6}= \end{equation} \begin{equation} =\frac{11 \cdot \sqrt[6]{2^{30}}}{6}= \end{equation} \begin{equation} =\frac{11 \cdot 2^{5}}{6}=\frac{11 \cdot 32}{6}= \end{equation} \begin{equation} =\frac{352}{6}=\frac{176}{3}. \end{equation}