Question

Товарищи,прошу вашей помощи,дуб-дубом))​

Answer 1

1) y = (2x² + 5x - 4)⁹

y' = 9(2x² + 5x - 4)⁸ * (2x² + 5x - 4)' = 9(2x² + 5x - 4)⁸ * (4x + 5) =

= 9(4x + 5)(2x² + 5x - 4)⁸

$2)y=\sqrt{2x^{2}-3x+7 }\\\\y'=\frac{1}{2\sqrt{2x^{2}-3x+7 } }*(2x^{2}-3x+7)'=\frac{4x-3}{2\sqrt{2x^{2}-3x+7 } }$

3) y = Sin3x

y' = Cos3x * (3x )' = 3Cos3x

4) y = Cos⁷x

y' = 7Cos⁶x * (Cosx)' = - 7Cos⁶xSinx

$5)y=e^{x^{7}-4 }\\\\y'=e^{x^{7}-4 }*(x^{7}-4)'=7x^{6}e^{x^{7} -4}$

$6)y=ln(x^{3}+6)\\\\y'=\frac{1}{x^{3}+6 }*(x^{3}+6)'=\frac{3x^{2} }{x^{3}+6 }$

$7)y=\frac{1}{(x^{2} +4)^{3} }=(x^{2}+4)^{-3}\\\\y'=-3(x^{2}+4)^{-4}*(x^{2}+4)'=-\frac{3}{(x^{2}+4)^{4}}*2x=-\frac{6x}{(x^{2}+4)^{4}  }$

$8)y=lnSin2x\\\\y'=\frac{1}{Sin2x}*(Sin2x)'=\frac{1}{Sin2x}*Cos2x*(2x)'=\frac{2Cos2x}{Sin2x}=2Ctg2x$

9)y = Sin²(3x + 5)

y'= 2Sin(3x+5)*(Sin(3x + 5))' = 2Sin(3x + 5)Cos(3x + 5)*(3x + 5)' =

= 6Sin(3x + 5)Cos(3x + 5) = 3Sin(6x + 10)

$10)y=x^{2}*lnx^{2} \\\\y'=(x^{2})'*lnx^{2}+x^{2}*lnx^{2} =2xlnx^{2} +x^{2}*\frac{1}{x^{2} }*(x^{2})'= 2xlnx^{2}+\frac{x^{2} *2x}{x^{2} } =2xlnx^{2}+2x=2x(lnx^{2}+1)$