вправа 20.6 гдз 10 клас математика Мерзляк Номіровський 2018
Вправа 20.6
Умова:
Умова:
Знайдіть похідну функції.
Відповідь ГДЗ:
\begin{equation} 1){y}'=\frac{{(3x+5)}'(x-8)-}{ \, } \end{equation} \begin{equation} \frac{-(3x+5){(x-8)}'}{(x-8)^{2}}= \end{equation} \begin{equation} =\frac{3(x-8)-(3x+5)}{(x-8)^{2}}= \end{equation} \begin{equation} =\frac{3x-24-3x+5}{(x-8)^{2}}= \end{equation} \begin{equation} =-\frac{19}{(x-8)^{2}}; \end{equation} \begin{equation} 2){y}'= \end{equation} \begin{equation} =\frac{{7}'(10x-3)-7{(10x-3)}'}{(10x-3)^{2}}= \end{equation} \begin{equation} =\frac{0-70}{(10x-3)^{2}}= \end{equation} \begin{equation} =-\frac{0-70}{(10x-3)^{2}}; \end{equation} \begin{equation} 3){y}'= \end{equation} \begin{equation} =\frac{{(2x^{2})}'(1-6x)-2x^{2}{(1-6x)}'}{(1-6x)^{2}}= \end{equation} \begin{equation} =\frac{4x(1-6x)-2x^{2} \cdot (-6)}{(1-6x)^{2}}= \end{equation} \begin{equation} =\frac{4x-24x^{2}+12x^{2}}{(1-6x)^{2}}= \end{equation} \begin{equation} =\frac{4x-12x^{2}}{(1-6x)^{2}}; \end{equation} \begin{equation} 4){y}'= \end{equation} \begin{equation} =\frac{{(\sin x)}' \cdot x-\sin x \cdot {x}'}{x^{2}}= \end{equation} \begin{equation} =\frac{x \, \cos x-\sin x}{x^{2}}; \end{equation} \begin{equation} 5){y}'=\frac{{(x^{2}-1)}'(x^{2}+1)-}{ \, } \end{equation} \begin{equation} \frac{-(x^{2}-1){(x^{2}+1)}'}{(x^{2}+1)^{2}}= \end{equation} \begin{equation} =\frac{2x(x^{2}+1)-(x^{2}-1) \cdot 2x}{(x^{2}+1)^{2}}= \end{equation} \begin{equation} =\frac{2x^{3}+2x-2x^{3}+2x}{(x^{2}+1)^{2}}= \end{equation} \begin{equation} =\frac{4x}{(x^{2}+1)^{2}}; \end{equation} \begin{equation} 6){y}'=\frac{{(x^{2}+6x)}'(x+2)-}{ \, } \end{equation} \begin{equation} \frac{-(x^{2}+6x){(x+2)}'}{(x+2)^{2}}= \end{equation} \begin{equation} =\frac{(2x+6)(x+2)-x^{3}-6x}{(x+2)^{2}}= \end{equation} \begin{equation} =\frac{2x^{2}+4x+6x+12-x^{2}-6x}{(x+2)^{2}}= \end{equation} \begin{equation} =\frac{x^{2}+4x+12}{(x+2)^{2}}. \end{equation}
\begin{equation} 1){y}'=\frac{{(3x+5)}'(x-8)-}{ \, } \end{equation} \begin{equation} \frac{-(3x+5){(x-8)}'}{(x-8)^{2}}= \end{equation} \begin{equation} =\frac{3(x-8)-(3x+5)}{(x-8)^{2}}= \end{equation} \begin{equation} =\frac{3x-24-3x+5}{(x-8)^{2}}= \end{equation} \begin{equation} =-\frac{19}{(x-8)^{2}}; \end{equation} \begin{equation} 2){y}'= \end{equation} \begin{equation} =\frac{{7}'(10x-3)-7{(10x-3)}'}{(10x-3)^{2}}= \end{equation} \begin{equation} =\frac{0-70}{(10x-3)^{2}}= \end{equation} \begin{equation} =-\frac{0-70}{(10x-3)^{2}}; \end{equation} \begin{equation} 3){y}'= \end{equation} \begin{equation} =\frac{{(2x^{2})}'(1-6x)-2x^{2}{(1-6x)}'}{(1-6x)^{2}}= \end{equation} \begin{equation} =\frac{4x(1-6x)-2x^{2} \cdot (-6)}{(1-6x)^{2}}= \end{equation} \begin{equation} =\frac{4x-24x^{2}+12x^{2}}{(1-6x)^{2}}= \end{equation} \begin{equation} =\frac{4x-12x^{2}}{(1-6x)^{2}}; \end{equation} \begin{equation} 4){y}'= \end{equation} \begin{equation} =\frac{{(\sin x)}' \cdot x-\sin x \cdot {x}'}{x^{2}}= \end{equation} \begin{equation} =\frac{x \, \cos x-\sin x}{x^{2}}; \end{equation} \begin{equation} 5){y}'=\frac{{(x^{2}-1)}'(x^{2}+1)-}{ \, } \end{equation} \begin{equation} \frac{-(x^{2}-1){(x^{2}+1)}'}{(x^{2}+1)^{2}}= \end{equation} \begin{equation} =\frac{2x(x^{2}+1)-(x^{2}-1) \cdot 2x}{(x^{2}+1)^{2}}= \end{equation} \begin{equation} =\frac{2x^{3}+2x-2x^{3}+2x}{(x^{2}+1)^{2}}= \end{equation} \begin{equation} =\frac{4x}{(x^{2}+1)^{2}}; \end{equation} \begin{equation} 6){y}'=\frac{{(x^{2}+6x)}'(x+2)-}{ \, } \end{equation} \begin{equation} \frac{-(x^{2}+6x){(x+2)}'}{(x+2)^{2}}= \end{equation} \begin{equation} =\frac{(2x+6)(x+2)-x^{3}-6x}{(x+2)^{2}}= \end{equation} \begin{equation} =\frac{2x^{2}+4x+6x+12-x^{2}-6x}{(x+2)^{2}}= \end{equation} \begin{equation} =\frac{x^{2}+4x+12}{(x+2)^{2}}. \end{equation}