Question

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Answer 1

$1)\ \ y=\Big(2-\dfrac{x}{3}\Big)\Big(\dfrac{1}{3}+x^2\Big)\\\\y'=-\dfrac{1}{3}\, \Big(\dfrac{1}{3}+x^2\Big)+2x\, \Big(2-\dfrac{x}{3}\Big)=-\dfrac{1}{9}-\dfrac{x^2}{3}+4x-\dfrac{2x^2}{3}=-x^2+4x-\dfrac{1}{9}\\\\\\2)\ \ y=\dfrac{x^2+3x+1}{x}=x+3+\dfrac{1}{x}\\\\y'=1-\dfrac{1}{x^2}\\\\\\3)\ \ y=e^{x^2}\ \ ,\ \ y'=e^{x^2}\cdot (x^2)'=2x\cdot e^{x^2}\\\\\\4)\ \ y=ln(sinx)\ \ ,\ \ y'=\dfrac{1}{sinx}\cdot (sinx)'=\dfrac{cosx}{sinx}=ctgx$

$5)\ \ y=u^2+3\sqrt{u}-1\ \ ,\ \ u=x^4+1\\\\y'=2u\cdot u'+\dfrac{3}{2\sqrt{u}}\cdot u'\\\\ili:\ \ \ y=(x^4+1)^2+3\sqrt{x^4+1}-1\\\\y'=2(x^4+1)\cdot 4x^3+\dfrac{3\cdot 4x^3}{2\sqrt{x^4+1}}=8x^3\cdot (x^4+1)+\dfrac{6x^3}{\sqrt{x^4+1}}$