Question

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

Answer 1

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

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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$\int \frac 1{cosx-1} dx$

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

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

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

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