Question

найти производную:

1) y= (x^2-3x+3)*(x^2+2x-1)
2)y=x^2+1/x^2-1
3)y= x*10^x
4) y=1-lnx/ 1+lnx
5) y=arctgx-arcctgx
6)y=(1+2x)^30 
7) y=(1+x^2/1+x)^5

8) у=sin^3x

9) y=ln cosx 

Answer 1

1) $y = (x^2-3x+3)(x^2+2x-1)$

$y' = (2x-3)(x^2+2x-1)+(x^2-3x+3)(2x+2) = 4x^3-3x^2-8x+9$ 

2) $y = x^2+frac{1}{x^2-1}$ 

$y' = 2x-frac{2x}{(x^2-1)^2} = 2x(1-frac{1}{(x^2-1)^2})$ 

3) $y=x*10^x$ 

$y' = 10^x+x*10^x*ln10 = 10^x(1+xln10)$ 

4) $y = frac{1-lnx}{1+lnx}$ 

$y' = frac{(1-lnx)'(1+lnx)-(1-lnx)(1+lnx)'}{(1+lnx)^2}=frac{-frac{1}{x}(1+lnx)-(1-lnx)frac{1}{x}}{(1+lnx)^2}=-frac{frac{1+lnx+1-lnx}{x}}{(1+lnx)^2}=-frac{2}{x(1+lnx)^2}$ 

5) y = arctgx - arcctgx

$y' = frac{1}{x^2+1}+frac{1}{x^2+1}=frac{2}{x^2+1}$ 

6) $y = (1+2x)^{30}$ 

$y' = 30(1+2x)^{29}*2 = 60(1+2x)^{29}$ 

7) $y = (1+frac{x^2}{1+x})^5$ 

$y' = 5(1+frac{x^2}{x+1})^4*frac{2x(1+x)-x^2}{(1+x)^2}=5frac{(x^2+x+1)^4}{(x+1)^4}*frac{2x+x^2}{(1+x)^2}=5xfrac{(x^2+x+1)^4(x+2)}{(x+1)^6}$

8) $y = sin^3x$ 

$y' = 3sin^2x*cosx$ 

9) y = lncosx

$y' = frac{1}{cosx}*(-sinx) = -tgx$