Question

Va rog ajutatima plsssdd

Answer 1

$\displaystyle \mathtt{a)~f(x)=2x^2+5x}\\ \\ \mathtt{f'(x)=\left(2x^2+5x\right)'=\left(2x^2\right)'+(5x)'=2\left(x^2\right)'+5x'=}\\ \\ \mathtt{=2 \cdot 2x+5\cdot1=4x+5}\\ \\ \mathtt{f''(x)=(f'(x))'=(4x+5)'=(4x)'+5'=4x'+0=4\cdot 1=4}$

$\displaystyle \mathtt{b)~f(x)=x^3-4x}\\ \\ \mathtt{f'(x)=\left(x^3-4x\right)'=\left(x^3\right)'-(4x)'=3x^2-4x'=3x^2-4\cdot1=3x^2-4}\\ \\ \mathtt{f''(x)=(f'(x))'=\left(3x^2-4\right)=\left(3x^2\right)-4'=3\left(x^2\right)-0=3 \cdot 2x=6x}$

$\displaystyle \mathtt{c)~f(x)=e^x+x}\\ \\ \mathtt{f'(x)=\left(e^x+x\right)'=\left(e^x\right)'+x'=e^x+1}\\ \\ \mathtt{f''(x)=(f'(x))'=\left(e^x+1\right)=\left(e^x\right)'+1'=e^x+0=e^x}$

$\displaystyle \mathtt{d)~f(x)=x+ln~x}\\ \\ \mathtt{f'(x)=(x+ln~x)'=x'+(ln~x)'=1+ \frac{1}{x}}\\ \\ \mathtt{f''(x)=(f'(x))'=\left(1+ \frac{1}{x}\right)'=1'+\left( \frac{1}{x}\right)'=0+\left(x^{-1}\right)'=-1 \cdot x^{-2}=}\\ \\ \mathtt{=- \frac{1}{x^2} }$

$\displaystyle \mathtt{e)~f(x)=xln~x}\\ \\ \mathtt{f'(x)=(xln~x)'=x' \cdot ln~x+x \cdot (ln~x)'=1 \cdot ln~x+x \cdot \frac{1}{x}=ln~x+1}\\ \\ \mathtt{f''(x)=(f'(x))'=(ln~x+1)'=(ln~x)'+1'= \frac{1}{x}+0= \frac{1}{x} }$

$\displaystyle \mathtt{f)~f(x)=x^2e^x}\\ \\ \mathtt{f'(x)=\left(x^2e^x\right)'=\left(x^2\right)' \cdot e^x+x^2 \cdot \left(e^x\right)'=2xe^x+x^2e^x}\\ \\ \mathtt{f''(x)=(f'(x))'=\left(2xe^x\right)'+\left(x^2e^x\right)'=2\left(xe^x\right)'+\left(x^2e^x\right)'=}\\ \\ \mathtt{=2\left(x' \cdot e^x+x \cdot \left(e^x\right)'\right)+\left(x^2\right)'\cdot e^x+x^2\cdot\left(e^x\right)'=}$

$\displaystyle \mathtt{=2\left(1 \cdot e^x+x \cdot e^x\right)+2xe^x+x^2e^x=2\left(e^x+xe^x\right)+2xe^x+x^2e^x=}\\ \\ \mathtt{=2e^x+2xe^x+2xe^x+x^2e^x=2e^x+4xe^x+x^2e^x}$