вправа 20.1 гдз 10 клас математика Мерзляк Номіровський 2018
Вправа 20.1
Умова:
Умова:
Знайдіть похідну функції.
Відповідь ГДЗ:
\begin{equation} 1){y}'={(x^{3})}'-{(3x^{2})}'+ \end{equation} \begin{equation} +{(6x)}'-{(10)}'= \end{equation} \begin{equation} =3x^{2}-3 \cdot 2x+6= \end{equation} \begin{equation} =3x^{2}-6x+6; \end{equation} \begin{equation} 2){y}'={(4x^{6})}'+{(20\sqrt{x})}'= \end{equation} \begin{equation} =4 \cdot 6x^{5}+20 \cdot \frac{1}{2\sqrt{x}}= \end{equation} \begin{equation} =24x^{5}+\frac{10}{\sqrt{x}}; \end{equation} \begin{equation} 3){y}'={(x^{3})}'+{(7x^{6})}'+ \end{equation} \begin{equation} +{\left ( \frac{4}{x} \right )}'-{1}'= \end{equation} \begin{equation} =8x^{7}+7 \cdot 6x+4 \cdot \left ( -\frac{1}{x^{2}} \right )= \end{equation} \begin{equation} =8x^{7}+42x-\frac{4}{x^{2}}; \end{equation} \begin{equation} 4){y}'={(4 \sin x)}'-{(5 \cos x)}'= \end{equation} \begin{equation} =4 \cos x+5 \sin x; \end{equation} \begin{equation} 5){y}'={(tg \, x)}'-{(9x)}'= \end{equation} \begin{equation} =\frac{1}{\cos^{2} x}-9; \end{equation} \begin{equation} 6){y}'={(2x^{-2})}'+{(3x^{-8})}'= \end{equation} \begin{equation} =2 \cdot {(-2x^{-3})}'+3 \cdot {(-3x^{-4})}'= \end{equation} \begin{equation} =-4x^{-3}-9x^{-4}. \end{equation}
\begin{equation} 1){y}'={(x^{3})}'-{(3x^{2})}'+ \end{equation} \begin{equation} +{(6x)}'-{(10)}'= \end{equation} \begin{equation} =3x^{2}-3 \cdot 2x+6= \end{equation} \begin{equation} =3x^{2}-6x+6; \end{equation} \begin{equation} 2){y}'={(4x^{6})}'+{(20\sqrt{x})}'= \end{equation} \begin{equation} =4 \cdot 6x^{5}+20 \cdot \frac{1}{2\sqrt{x}}= \end{equation} \begin{equation} =24x^{5}+\frac{10}{\sqrt{x}}; \end{equation} \begin{equation} 3){y}'={(x^{3})}'+{(7x^{6})}'+ \end{equation} \begin{equation} +{\left ( \frac{4}{x} \right )}'-{1}'= \end{equation} \begin{equation} =8x^{7}+7 \cdot 6x+4 \cdot \left ( -\frac{1}{x^{2}} \right )= \end{equation} \begin{equation} =8x^{7}+42x-\frac{4}{x^{2}}; \end{equation} \begin{equation} 4){y}'={(4 \sin x)}'-{(5 \cos x)}'= \end{equation} \begin{equation} =4 \cos x+5 \sin x; \end{equation} \begin{equation} 5){y}'={(tg \, x)}'-{(9x)}'= \end{equation} \begin{equation} =\frac{1}{\cos^{2} x}-9; \end{equation} \begin{equation} 6){y}'={(2x^{-2})}'+{(3x^{-8})}'= \end{equation} \begin{equation} =2 \cdot {(-2x^{-3})}'+3 \cdot {(-3x^{-4})}'= \end{equation} \begin{equation} =-4x^{-3}-9x^{-4}. \end{equation}