Question

У'=???? ответы?? 787-820

Answer 1

800.
$y = 1- e^{sin^2 3x}cos^2 3x$
$y' = 0-[(e^{sin^2 3x})'*cos^2 3x+e^{sin^2 3x}*(cos^2 3x)'] = \\ \\ = -[e^{sin^2 3x} *(sin^2 3x)' *cos^2 3x+ \\ \\ + e^{sin^2 3x} *2cos3x *(cos3x)'] = \\ \\ = -[e^{sin^2 3x} *2sin3x *(sin3x)' *cos^2 3x+ \\ \\ + e^{sin^2 3x} *2cos3x *(-3sin3x) ] = \\ \\ = - 6e^{sin^2 3x} [sin3x *cos3x *cos^2 3x - cos3x *sin3x] =$
$= - 6e^{sin^2 3x}*sin3x *cos3x* [cos^2 3x - 1] = \\ \\ = - 6e^{sin^2 3x}*sin3x *cos3x* [-sin^2 3x] = 6e^{sin^2 3x}*sin^33x *cos3x$

801.
$y = ln \frac{2ln^2 sinx + 3}{2ln^2 sinx - 3} \\ \\ y' = \frac{1}{ \frac{2ln^2 sinx + 3}{2ln^2 sinx - 3} } *( \frac{2ln^2 sinx + 3}{2ln^2 sinx - 3} )'= \\ \\ = \frac{2ln^2 sinx - 3}{2ln^2 sinx + 3} * \frac{(2ln^2 sinx + 3)'*(2ln^2 sinx - 3) - (2ln^2 sinx + 3)*(2ln^2 sinx - 3)' }{(2ln^2 sinx - 3)^2} = \\ \\ = \frac{1}{4ln^4 sinx - 9} * [4lnsinx* \frac{1}{sinx}*cosx* (2ln^2 sinx - 3) - \\ \\ - (2ln^2 sinx + 3)*4lnsinx * \frac{1}{sinx} *cosx ]=$
$= \frac{4lnsinx*ctgx}{4ln^4 sinx - 9} *[2ln^2 sinx - 3 -(2ln^2 sinx + 3) ] = \\ \\ =\frac{4lnsinx*ctgx}{4ln^4 sinx - 9} * (-6) = -24\frac{lnsinx*ctgx}{4ln^4 sinx - 9}$