Question

Найти производные y=ln^2 arctg x:3 ;y =(2x+1)^3x ;y=x^3(x+2)^2

Answer 1

$\mathtt{f(x)=ln^2arctg(\frac{x}{3});~f'(x)=2lnarctg(\frac{x}{3})[lnarctg(\frac{x}{3})]'=}\\\mathtt{2lnarctg(\frac{x}{3})*\frac{1}{arctg(\frac{x}{3})}[arctg(\frac{x}{3})]'=\frac{2lnarctg(\frac{x}{3})}{arctg(\frac{x}{3})}*\frac{1}{1+(\frac{x}{3})^2}*(\frac{x}{3})'=}\\\mathtt{2log_{e^{arctg(\frac{x}{3})}}arctg(\frac{x}{3}):\frac{x^2+9}{9}*\frac{1}{3}=\frac{6lnarctg(\frac{x}{3})}{(x^2+9)arctg(\frac{x}{3})}}$

ОТВЕТ: $\mathtt{f'(ln^2arctg(\frac{x}{3}))=\frac{6lnarctg(\frac{x}{3})}{(x^2+9)arctg(\frac{x}{3})}}$


$\mathtt{f(x)=(2x+1)^{3x};~f'(x)=(2x+1)^{3x}ln(2x+1)*(3x)'*[(2x+1)^3]'=}\\\mathtt{9ln(2x+1)^{(2x+1)^{3x+2}}}$

ОТВЕТ: $\displaystyle\mathtt{f'((2x+1)^{3x})=9ln(2x+1)^{(2x+1)^{3x+2}}}$


$\mathtt{f(x)=x^5+4x^4+4x^3;~f'(x)=(x^5+4x^4+4x^3)'=5x^4+16x^3+12x^2}$

ОТВЕТ: $\mathtt{f'(x^3(x+2)^2)=5x^4+16x^3+12x^2}$