Question

найти производную функции: y=(tgx-sinx)/x^3 найти первообразную функции y=1/(cosx-1)

Answer 1

$\y=frac{tan x-sin x}{x^3}\ y'=frac{(sec^2x-cos x)cdot x^3-(tan x-sin x)cdot3x^2}{x^6}\ y'=frac{x^2(x(sec^2x-cos x)-3tan x+3sin x)}{x^6}\ y'=frac{xsec^2x-xcos x-3tan x+3sin x}{x^4}$

Answer 2

$f(x) = frac{tgx-sinx}{x^3}$

 

$f(x) = frac{g(x)}{h(x)}$    

 

$frac {d}{dx} frac {g(x)}{h(x)} = frac {frac d{dx} g(x) cdot h(x) - frac d{dx} h(x) cdot g(x)}{(h(x))^2}$

 

$g(x) ={tgx-sinx}$     $h(x)={x^3}$

 

$frac d{dx} g(x) =frac d{dx}tgx- frac d{dx}sinx = frac 1 {cos^2 x} - cosx$

 

$frac d{dx} h(x) = frac d{dx} x^3 =3x^2$

 

$frac {d}{dx} f(x) = frac {(frac 1 {cos^2 x} - cosx) cdot x^3 - 3x^2 cdot (tgx - sinx))}{(x^3)^2} =$

 

$frac {x^3 (frac 1 {cos^2 x} - cosx) - 3x^2 cdot (tgx - sinx))}{x^6} =$

 

$frac {x (frac 1 {cos^2 x} - cosx) - 3 (tgx - sinx))}{x^4}$

 

если ползуете функцей секанс :   $frac 1 {cos^2 x} = sec^2 x$

 

$frac {x (sec^2 x - cosx) - 3 (tgx - sinx))}{x^4}$

 

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$int frac 1{cosx-1} dx$

 

$u =tg frac x2$               $du = frac {sec^2(frac x2) dx }{2}$

 

$sin x = frac {2u}{1 +u^2}$                $cos x =frac {1-u^2}{1+u^2}$

 

$int frac 2{ (1+u^2)(-1+frac {1-u^2}{1+u^2})} du = int -frac 1{u^2} du= frac1 u + C=$

 

 $frac1 {tg frac x2} + C = ctg frac x 2 +C$