Установить соответствие между функциями и их производными.

Ответ:

1-В, 2-А, 3-Д, 4-Г.

Объяснение:

1) y=4x⁴-(2/x²)

[tex]\LARGE \boldsymbol {} y= 4x^4-\frac{2}{x^2} \\\\y'=(4x^4-\frac{2}{x^2})'=(4x^4)'-(\frac{2}{x^2})'=4*4x^{4-1} -\\\\-\frac{(2)'x^2-(x^2)'*2}{(x^2)^2} =16x^3-\frac{0*x^2-2x^{2-1}*2 }{x^4} =\\\\=16x^3-\frac{-4x}{x^4} =16x^3+\frac{4}{x^3}[/tex]

1-В: y=4x⁴-(2/x²) ⇒ y'=16x³+(4/x³)

2) y=(x⁴/4)-(1/2x²)

[tex]\LARGE \boldsymbol {} y=\frac{x^4}{4}-\frac{1}{2x^2} \\\\y'=(\frac{x^4}{4}-\frac{1}{2x^2} )'=(\frac{1}{4} x^4)'-(\frac{1}{2x^2} )'=\\\\=\frac{1}{4} *4x^{4-1} -\frac{(1)'*2x^2-(2x^2)'*1}{(2x^2)^2} =x^{3} -\\\\-\frac{0*2x^2-2*2x^{2-1} }{4x^4} =x^3-\frac{-\not4x}{\not4x^4} =x^3+\frac{1}{x^3}[/tex]

2-A: y=(x⁴/4)-(1/2x²) ⇒ y'=x³+(1/x³)

3)  y=(x⁴/4)-(2/x²)

[tex]\LARGE \boldsymbol {} y=\frac{x^4}{4}-\frac{2}{x^2} \\\\y'=(\frac{x^4}{4}-\frac{2}{x^2})'=(\frac{1}{4}x^4)'-(\frac{2}{x^2})'=\frac{1}{\not4}*\not4x^{4-1} -\\\\-\frac{(2)'x^2-(x^2)'*2}{(x^2)^2}=x^3-\frac{0*x^2-2x^{2-1}*2 }{x^4} =\\\\=x^3-\frac{-4x}{x^4} =x^3+\frac{4}{x^3}[/tex]

3-Д: y=(x⁴/4)-(2/x²) ⇒ y'=x³+(4/x³)

4) y=(4x⁴)-(1/2x²)

[tex]\LARGE \boldsymbol {} y=4x^4-\frac{1}{2x^2} \\\\y'=(4x^4-\frac{1}{2x^2} )'=(4x^4)'-(\frac{1}{2x^2} )'=4*4x^{4-1}-\\\\-\frac{(1)'*2x^2-(2x^2)'*1}{(2x^2)^2} =16x^3-\frac{0*2x^2-2*2x^{2-1} }{4x^4}=\\\\=16x^3-\frac{-4x}{4x^4}=16x^3+\frac{1}{x^3}[/tex]

4-Г: y=(4x⁴)-(1/2x²) ⇒ y=16x³+(1/x³)