1)f(x)=\sqrt(6x-7) при x(0)=3
2)f(x)=cos^4x при x(0)=pi/4
найти производную данных функций и вычислить их значение в данной точке х(0)
1) f ' (x) = (√ (6 * x - 7))' = (6 * x - 7)'/(2 * √ (6 * x - 7)) = 6 /(2 * √ (6 * x - 7)) =
3 / √ (6 * x - 7)
При Х0 = 3 f ' (3) = 3 / √ (6 * 3 - 7) = 3 / √ 11.
'2) f ' (x) = (cos⁴x)' = 4 * cos³x * (cos x)' = 4 * cos³x * (- sin x) = -4 * cos³x * sin x
При Х0 = π/4 f ' (π/4) = -4 * cos³(π/4) * sin π/4 = -4 * (1 / √ 2)³ * (1/ √ 2) = -1
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1) f ' (x) = (√ (6 * x - 7))' = (6 * x - 7)'/(2 * √ (6 * x - 7)) = 6 /(2 * √ (6 * x - 7)) =
3 / √ (6 * x - 7)
При Х0 = 3 f ' (3) = 3 / √ (6 * 3 - 7) = 3 / √ 11.
'2) f ' (x) = (cos⁴x)' = 4 * cos³x * (cos x)' = 4 * cos³x * (- sin x) = -4 * cos³x * sin x
При Х0 = π/4 f ' (π/4) = -4 * cos³(π/4) * sin π/4 = -4 * (1 / √ 2)³ * (1/ √ 2) = -1