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) $y = (1+\frac{x^2}{1+x})^5$ 

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

8) $y = sin^3x$ 

$y' = 3sin^2x*cosx$ 

9) y = lncosx

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