Question

СРОЧНО. Найти 7 производных. С решением пожалуйста.

Answer 1

Объяснение:

1)

$\displaystyle y=-4ctg^3x\\\\y'=-4*3ctg^2x*(ctgx)'=-12ctg^2x*(-\frac{1}{sin^2x})=\frac{12ctg^2x}{sin^2x}$

2)

$\displaystyle y=-\frac{1}{5}cos^5x\\\\y'=-\frac{1}{5}*5cos^4x*(cosx)'=-cos^4x*(-sinx)=cos^4x*sinx$

3)

$\displaystyle y=\frac{1}{lnx}\\\\y'=-\frac{1}{ln^2x}*(lnx)'=-\frac{1}{ln^2x}*\frac{1}{x}=-\frac{1}{x\;ln^2x}$

4)

$\displaystyle y=\sqrt{(2-x)(3-2x)}=\sqrt{6-4x-3x+2x^2}=({6-7x+2x^2} )^{\frac{1}{2} }\\\\y'=\frac{1}{2}(6-7x+2x^2)^{-\frac{1}{2} }*(6-7x+2x^2)'=\frac{-7+2*2x}{2\sqrt{6-7x+2x^2} } = \\\\= \frac{4x-7}{2\sqrt{6-7x+2x^2} }$

5)

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

6)

$\displaystyle y=\sqrt[5]{\frac{1}{\sqrt{x} +1} } =(\frac{1}{\sqrt{x} +1} )^{\frac{1}{5} }\\\\y'=\frac{1}{5}(\frac{1}{\sqrt{x} +1} )^{-\frac{4}{5} } *(\frac{1}{\sqrt{x} +1} )'=\frac{1}{5}(\sqrt{x} +1)^{\frac{4}{5} }*(-\frac{1}{(\sqrt{x} +1)^2})*(\sqrt{x} +1)'=\\\\=\frac{1}{5}\sqrt[5]{(\sqrt{x} +1)^4}*(-\frac{1}{(\sqrt{x} +1)^2})*\frac{1}{2}x^{-\frac{1}{2} }=\\\\=-\frac{1}{10}\sqrt[5]{(\sqrt{x} +1)^4} *\frac{1}{\sqrt{x} (\sqrt{x} +1)^2}$

7)

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