1) Д
[tex]y = x \sqrt{2} \\ y' = \sqrt{2} \\ y'(2) = \sqrt{2} [/tex]
2) Б
[tex]y = \sqrt{2x} \\ y' = \frac{2}{2 \sqrt{2x} } = \frac{1}{ \sqrt{2x} } \\ y'(2) = \frac{1}{ \sqrt{2 \times 2} } = \frac{ 1 }{ \sqrt{ {2}^{2} } } = \frac{1}{2} [/tex]
3) А
[tex]y = \sqrt{x + 2} \\ y' = \frac{1}{2 \sqrt{x + 2} } \\ y'(2) = \frac{1}{2 \sqrt{2 + 2} } = \frac{1}{2 \sqrt{4} } = \frac{1}{2 \times 2} = \frac{1}{4} [/tex]
4) В
[tex]y = \sqrt{2x - 3} \\ y' = \frac{2}{2 \sqrt{2x - 3} } = \frac{1}{ \sqrt{2x - 3} } \\ y' (2)= \frac{1}{ \sqrt{2 \times 2 - 3} } = \frac{1}{ \sqrt{4 - 3} } = \frac{1}{ \sqrt{1} } = 1[/tex]
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Answers & Comments
1) Д
[tex]y = x \sqrt{2} \\ y' = \sqrt{2} \\ y'(2) = \sqrt{2} [/tex]
2) Б
[tex]y = \sqrt{2x} \\ y' = \frac{2}{2 \sqrt{2x} } = \frac{1}{ \sqrt{2x} } \\ y'(2) = \frac{1}{ \sqrt{2 \times 2} } = \frac{ 1 }{ \sqrt{ {2}^{2} } } = \frac{1}{2} [/tex]
3) А
[tex]y = \sqrt{x + 2} \\ y' = \frac{1}{2 \sqrt{x + 2} } \\ y'(2) = \frac{1}{2 \sqrt{2 + 2} } = \frac{1}{2 \sqrt{4} } = \frac{1}{2 \times 2} = \frac{1}{4} [/tex]
4) В
[tex]y = \sqrt{2x - 3} \\ y' = \frac{2}{2 \sqrt{2x - 3} } = \frac{1}{ \sqrt{2x - 3} } \\ y' (2)= \frac{1}{ \sqrt{2 \times 2 - 3} } = \frac{1}{ \sqrt{4 - 3} } = \frac{1}{ \sqrt{1} } = 1[/tex]