Question

Срочно!!! Найти производные функций

Answer 1

$1.y' =2sinxcosx*lnx+ \frac{sin^2x}{x} =sin2x*lnx+ \frac{sin^2x}{x} \\\\ 2.y'= \frac{(2x+2)(3-4x)-(x^2+2x)(-4)}{(3-4x)^2} = \frac{-2x-8x^2+6+4x^2+8x}{(3-4x)^2} = \frac{-4x^2+6x+6}{(3-4x)^2} \\\\ 3.y'= \frac{(ctg^3x)'}{ctg^3x} =- \frac{3ctg^2x}{sin^2xctg^3x} =- \frac{3}{sinxcosx} =- \frac{6}{sin2x} \\ y'( \frac{ \pi }{4} )=- \frac{6}{sin( \frac{\pi}{2} } =-6$

$5. log_5y=3x^3+x\\\\ \frac{y'}{y*ln5} =9x^2+1\\\\ y'=5^{3x^3+x}(9x^2+1)ln5\\\\ 7.y'= \frac{( \frac{x-1}{x+1} )'}{1+( \frac{x-1}{x+1} )^2} = \frac{ \frac{2}{(x+1)^2} }{ \frac{x^2+2x+1+x^2-2x+1}{(x+1)^2} } = \frac{1}{x^2+1}$

Answer 2

1. $y'=(sin^{2}(x))'*lnx+sin^{2}(x)*ln'x=2sin(x)cos(x)ln(x) - \frac{sin(x)^{2}}{x}$
2. $y'= \frac{( x^{2} +2x)'(3-4x)-( x^{2} +2x)(3-4x)'}{(3-4x)^{2}} =\frac{-4 x^{2} +6x+6}{ (3-4x)^{2} }$
3. $y' = \frac{1}{ctg^{3}x}*3ctg^{2}x* \frac{1}{sin^{2}x} = \frac{-3}{cos(x)sin(x)}$   $y'( \frac{ \pi}{4} ) = \frac{-3}{ \frac{1}{2}} = -6$  
5. $y'=5^{3x^{3} -x} (9 x^{2}-1 )ln(5)$
7. $y'= \frac{1}{ ( \frac{x-1}{x+1} )^{2} +1}*(\frac{x-1}{x+1})'= \frac{ \frac{2}{(x+1)^{2}}}{ \frac{ x^{2} +2x+1+ x^{2} x-2x+1}{(x+1)^{2}}} =\frac{1}{ x^{2}+1}$
все решается просто подставляя в формулы