Question

вычислите y' в точке x0:
y= 5x-3/3x-3, x0=5​

Answer 1

$y = \frac{5x - 3}{3x - 3} \\$

$y '= \frac{(5x - 3)'(3x - 3) - (3x - 3)'(5x - 3)}{ {(3x - 3)}^{2} } = \\ = \frac{5(3x - 3) - 3(5x - 3)}{ {(3x - 3)}^{2} } = \frac{15x - 15 - 15x + 9}{ {(3x - 3)}^{2} } = \\ = \frac{ - 6}{ {(3x - 3)}^{2} } = \frac{ - 6}{ {(3(x - 1))}^{2} } = \\ = \frac{ - 6}{9 {(x - 1)}^{2} } = \frac{ - 2}{3 {(x - 1)}^{2} }$

$y'(5) = \frac{ - 2}{3 \times {4}^{2} } = - \frac{1}{3 \times 8} = - \frac{1}{24} \\$

Answer 2

решение смотри на фотографии