Question

ajutaţi-mă vă rog! îmi trebuie până mâine

Answer 1

$\displaystyle \mathtt{a)~f(x)=x^3+3x+1}\\ \\ \mathtt{f'(x)=\left(x^3+3x+1\right)'=\left(x^3\right)'+(3x)'+1'=3x^2+3x'+0=}\\ \\ \mathtt{=3x^2+3\cdot1=3x^2+3}$

$\displaystyle \mathtt{b)~f(x)=2x-x^4}\\ \\ \mathtt{f'(x)=\left(2x-x^4\right)'=(2x)'-\left(x^4\right)'=2x'-4x^3=2\cdot1-4x^3=2-4x^3}$

$\displaystyle \mathtt{c)~f(x)=x+2 \sqrt{x} }\\ \\ \mathtt{f'(x)=\left(x+2 \sqrt{x} \right)'=x'+\left(2 \sqrt{x} \right)'=1+2\left( \sqrt{x} \right)'=1+\not2\cdot \frac{1}{\not2 \sqrt{x} } =}\\ \\ \mathtt{=1+ \frac{1}{ \sqrt{x} } }$

$\displaystyle \mathtt{d)~f(x)=x^3+sin~x+cos~x}\\ \\ \mathtt{f'(x)=\left(x^3\right)'+(sin~x)'+(cos~x)'=3x^2+cos~x-sin~x}$