Question

Продиференціювати функції.
Знайти похідну ....... та .......


Продифференцировать функции.

Найти производную ....... и .......

Answer 1

$\displaystyle 4)\\\\ y=\frac{x}{x^2+1}$

$\displaystyle y'=\left(\frac{x}{x^2+1}\right)'=\frac{x'(x^2+1)-(x^2+1)'x}{(x^2+1)^2}=$

$\displaystyle \frac{1\cdot (x^2+1)-2x\cdot x}{(x^2+1)^2}=\frac{x^2+1-2x^2}{(x^2+1)^2}=\frac{-x^2}{(x^2+1)^2}$

$\displaystyle y''=\left(\frac{-x^2}{(x^2+1)^2}\right)'=$

$\displaystyle \frac{(-x^2)'(x^2+1)^2-((x^2+1)^2)'(-x^2)}{((x^2+1)^2)^2}=$

$\displaystyle \frac{-2x(x^2+1)^2-2(x^2+1)\cdot(x^2+1)'(-x^2)}{(x^2+1)^4}=$

$\displaystyle \frac{-2x(x^2+1)^2+2x^2(x^2+1)\cdot 2x}{(x^2+1)^4}=$

$\displaystyle \frac{-2x(x^2+1)^2+4x^3(x^2+1)}{(x^2+1)^4}=$

$\displaystyle \frac{2x(x^2+1)(-x^2-1+2x^2)}{(x^2+1)^4}=\frac{2x(x^2+1)}{(x^2-1)^3}$

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$\displaystyle 5)\\\\ y=\frac{2x^2+5x-1}{2x^2+5x+4}$

$\displaystyle y'= \left(\frac{2x^2+5x-1}{2x^2+5x+4}\right)'=$

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

$\displaystyle \frac{(4x+5)(2x^2+5x+4)-(2x^2+5x-1)(4x+5)}{(2x^2+5x+4)^2}=$

$\displaystyle \frac{(4x+5)(2x^2+5x+4-2x^2-5x+1)}{(2x^2+5x+4)^2}=\frac{5(4x+5)}{(2x^2+5x+4)^2}$

$\displaystyle y''= \left(\frac{5(4x+5)}{(2x^2+5x+4)^2}\right) '=\left(\frac{20x+25}{(2x^2+5x+4)^2}\right) '=$

$\displaystyle \frac{(20x+25)'(2x^2+5x+4)^2-(20x+25)((2x^2+5x+4)^2)'}{(2x^2+5x+4)^4}=$

$\displaystyle \frac{20(2x^2+5x+4)^2-(20x+25)\cdot 2(2x^2+5x+4)\cdot(2x^2+5x+4)' }{(2x^2+5x+4)^4}=$

$\displaystyle \frac{20(2x^2+5x+4)^2-(20x+25)\cdot 2(2x^2+5x+4)\cdot(4x+5)}{(2x^2+5x+4)^4}=$

$\displaystyle \frac{(2x^2+5x+4)[20(2x^2+5x+4)-2(20x+25)(4x+5)]}{(2x^2+5x+4)^4}=$

$\displaystyle \frac{20(2x^2+5x+4)-2(20x+25)(4x+5)}{(2x^2+5x+4)^3}=$

$\displaystyle \frac{40x^2+100x+80-2\cdot 5(4x+5)(4x+5)}{(2x^2+5x+4)^3}=$

$\displaystyle \frac{40x^2+100x+80-10(16x^2+40x+25)}{(2x^2+5x+4)^3}=$

$\displaystyle \frac{40x^2+100x+80-160x^2-400x+250}{(2x^2+5x+4)^3}=\frac{- 120x^2 - 300x - 170}{(2x^2+5x+4)^3}$

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$\displaystyle 6)\\\\ y=4 sin(3x+5)$

$\displaystyle y'= \left(4 sin(3x+5)\right)'=4 (sin(3x+5))'=4 cos(3x+5)\cdot (3x+5)'=$

$\displaystyle 4 cos(3x+5)\cdot 3=12 cos(3x+5)$

$\displaystyle y''=\left(12 cos(3x+5)\right)'=12(cos(3x+5))'=$

$\displaystyle 12 (-sin(3x+5))\cdot (3x+5)'=-12 sin(3x+5)\cdot 3=-36sin(3x+5)$