Question

Вычислить производную сложной функции. у=√sin(2x²-π) ͞8

Answer 1

1/2*cos(x^2-П/8)*4x/sqrt(sin(2x^2-П/2))

Answer 2

$y'=(\sqrt {sin {\frac {2x^2-\pi}{8}}})'= \frac{(sin {\frac {2x^2-\pi}{8}})'} {2\sqrt {sin {\frac {2x^2-\pi}{8}}}}= \frac {cos {\frac {2x^2-\pi}{8}}} {2\sqrt {sin {\frac {2x^2-\pi}{8}}}} *(\frac {2x^2-\pi}{8})'}=$

$=\frac {cos {\frac {2x^2-\pi}{8}}} {2\sqrt {sin {\frac {2x^2-\pi}{8}}}} *(\frac {1}{8})}*(2x^2-\pi)'= =\frac {cos {\frac {2x^2-\pi}{8}}} {2\sqrt {sin {\frac {2x^2-\pi}{8}}}} *(\frac {1}{8})}*(4x-0)=$

$=\frac {x cos {\frac {2x^2-\pi}{8}}} {4\sqrt {sin {\frac {2x^2-\pi}{8}}}}$