Ответ:
y = 2cos3x
a) y' = 2 • 3 • (-sin3x) = -6sin3x.
b) y = f'(x0) • (x - x0) + f(x0).
f'(П/6) = -6 • sin (3 • П/6) = -6 sin (П/2) = -6
f(П/6) = 2 cos (3 • П/6) = 2 cos(П/2) = 0
y = -6 (x - П/6) + 0 = -6(x - П/6).
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Answers & Comments
Ответ:
y = 2cos3x
a) y' = 2 • 3 • (-sin3x) = -6sin3x.
b) y = f'(x0) • (x - x0) + f(x0).
f'(П/6) = -6 • sin (3 • П/6) = -6 sin (П/2) = -6
f(П/6) = 2 cos (3 • П/6) = 2 cos(П/2) = 0
y = -6 (x - П/6) + 0 = -6(x - П/6).