Question

Найдите производные следующие функции: 1) y=(3-2x^3 )^5 2) y=(x^4-x-1)^4 3) y= √(x^3+1) 4) y= ^3√((1-x^2 )^2 ) 5) y=2*cos5x 6) y=ln(sinx) 7) y=ln^3(x^2-1) 8) y = ln(ln^35x)

Answer 1

1)$(3-2x^{3})^{5}=5(3-2x^{3})^{4}((3-2x^{3}))'=-10(3-2x^{3})^{4}*( x^{3} )'$
$=-30x^{2}(3-2x^{3})^{4}$
2)$=4(x^{4}-x-1)^{3}*(x^{4}-x-1)'=4(x^{4}-x-1)^{3}*(4 x^{3} -1)$
3)$(\sqrt{x^{3}+1})'= \frac{3x^{2}}{2 \sqrt{x^{3}+1} }$
4)$= \frac{(1- x^{2} )^{2}'}{3((1- x^{2} )^{2})^ \frac{2}{3} } = \frac{2(1- x^{2} )*(1- x^{2} )'}{3((1- x^{2} )^{2})^ \frac{2}{3} } =$
$= \frac{4x(1- x^{2} )}{3((1- x^{2} )^{2})^ \frac{2}{3} }$
5)$(2cos5x)'=-10sin5x$
6)$ln(sinx)'=?$
7)$ln^{3}( x^{2} -1)'=3ln^{2}( x^{2} -1)*ln( x^{2} -1)'=$
$=3ln^{2}( x^{2} -1)* \frac{( x^{2} -1)'}{ x^{2} -1} =\frac{6x*ln^{2}( x^{2} -1) }{ x^{2} -1}$
8)$ln(ln^{35}x)'=35*ln(lnx)'=35 \frac{lnx'}{lnx} = \frac{1}{x} * \frac{35}{lnx}$