Question

СРОЧНО. Найти 7 производных. С решением, пожалуйста.

Answer 1

Ответ:

$1)\ \ y=-4ctg^3x\\\\y'=-4\cdot 3\, ctg^2x\cdot \dfrac{-1}{sin^2x}=12\cdot ctg^2x\cdot \dfrac{1}{sin^2x}\\\\\\2)\ \ y=-\dfrac{1}{5}\, cos^5x\\\\y'=-\dfrac{1}{5}\cdot 5\, cos^4x\cdot (-sinx)=cos^4x\cdot sinx\\\\\\3)\ \ y=\dfrac{1}{lnx}\\\\y'=-\dfrac{1}{ln^2x}\cdot \dfrac{1}{x}=-\dfrac{1}{x\cdot ln^2x}\\\\\\4)\ \ y=\sqrt{(2-x)(3-2x)}=\sqrt{6-7x+2x^2}\\\\y'=\dfrac{1}{2\sqrt{6-7x+2x^2}}\cdot (4x-7)$

$5)\ \ y=(x+1)\sqrt{x^2+1}\\\\y'=\sqrt{x^2+1}+(x+1)\cdot \dfrac{1}{2\sqrt{x^2+1}}\cdot 2x=\dfrac{x^2+1+x(x+1)}{\sqrt{x^2+1}}=\dfrac{2x^2+2}{\sqrt{x^2+1}}=\\\\=2\sqrt{x^2+1}\\\\\\6)\ \ y=\sqrt[5]{\dfrac{1}{\sqrt{x}+1}}$

$y'=\dfrac{1}{5}\cdot \Big(\dfrac{1}{\sqrt{x}+1}\Big)^{-\frac{4}{5}}\cdot \dfrac{-\dfrac{1}{2\sqrt{x}}}{(\sqrt{x}+1)^2}=\dfrac{1}{5}\cdot \Big(\dfrac{1}{\sqrt{x}+1}\Big)^{-\frac{4}{5}}\cdot \dfrac{1}{2\sqrt{x}\, (\sqrt{x}+1)}=\\\\=\dfrac{1}{5}\cdot \Big(\sqrt{x}+1\Big)^{\frac{4}{5}}\cdot \dfrac{1}{2\sqrt{x}\, (\sqrt{x}+1)}\\\\\\7)\ \ y=\sqrt{x-1}\cdot (x+1)^2\\\\y'=\dfrac{1}{2\sqrt{x-1}}\cdot (x+1)^2+\sqrt{x-1}\cdot 2(x+1)=\dfrac{(x+1)^2+4(x+1)(x-1)}{2\sqrt{x-1}}=\\\\=\dfrac{5x^2+2x-3}{2\sqrt{x-1}}$