производные СРОЧНО!,с решением. Created by lembright. algebra-ru

производные СРОЧНО!,с решением...

Ответ:

$1)\ \ y=x^2\ \ ,\ \ \ y'=2x\\\\2)\ \ y=sinx\ \ ,\ \ \ y'=cosx\\\\3)\ \ y=2^{x}\ \ ,\ \ \ y'=2^{x}\, ln2\\\\4)\ \ y=4x^3+11x+\dfrac{3}{x^5}\ \ ,\ \ y'=12x^2+11-\dfrac{15}{x^6}\\\\5)\ \ y=6\cdot 5^{x}+7\, e^{x}\ \ ,\ \ y'=6\, ln5\cdot 5^{x}+7\, e^{x}\\\\6)\ \ y=lnx+4x^6-cosx+8\ \ ,\ \ y'=\dfrac{1}{x}+24x^5+sinx\\\\7)\ \ y=5\, tgx-9\, sinx+x\ \ ,\ \ y'=\dfrac{5}{cos^2x}-9\, cosx+1\\\\8)\ \ y=7\, arcsinx-3\, arccosx\ \ ,\ \ \ y'=\dfrac{7}{\sqrt{1-x^2}}+\dfrac{3}{\sqrt{1-x^2}}=\dfrac{10}{\sqrt{1-x^2}}$

$9)\ \ y=x^6\cdot tgx\ \ ,\ \ \ y'=6x^5\cdot tgx+\dfrac{x^6}{cos^2x}\\\\10)\ \ y=(x+1)\, e^{x}\ \ ,\ \ \ y'=e^{x}+(x+1)\, e^{x}=e^{x}\, (x+2)\\\\11)\ \ y=\dfrac{cosx}{sinx}=ctgx\ \ ,\ \ \ y'=-\dfrac{1}{sin^2x}\\\\\star \ \ y=\dfrac{cosx}{lnx}\ \ ,\ \ y'= \dfrac{-sinx\cdot lnx-cosx\cdot \dfrac{1}{x}}{ln^2x}=\dfrac{-x\cdot sinx\cdot lnx-cosxx}{x\cdot ln^2x}$

$\12)\ \ y=\dfrac{x^2-3x+1}{arcsinx}\\\\y'=\dfrac{(2x-3)\, arcsinx-(x^2-3x+1)\cdot \dfrac{1}{\sqrt{1-x^2}} }{arcsin^2x}=\\\\=\dfrac{(2x-3)\sqrt{1-x^2}\, arcsinx-x^2+3x-1}{\sqrt{1-x^2}\, arcsin^2x }$

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