Question

f(x)= x log in baza 2 din x
f(x)= x sin x
f(x)= x^2 cos x
f(x)= 1/x
f(x)= x-1/x+1
f(x)= 1+e^x/2+e^x
Va roggg ma puteti ajuta cu aceste derivate??

Answer 1

$\displaystyle \mathtt{1)~f(x)=xlog_2x}\\ \\ \mathtt{f'(x)=(xlog_2x)'=x'\cdot log_2x+x \cdot (log_2x)'=1\cdot log_2x+x\cdot \frac{1}{xln~2}=}\\ \\ \mathtt{=log_2x+ \frac{1}{ln~2} }$

$\displaystyle \mathtt{2)~f(x)=xsin~x}\\ \\ \mathtt{f'(x)=(xsin~x)'=x'\cdot sin~x+x\cdot(sin~x)'=1\cdot sin~x+x \cdot cos~x=}\\ \\ \mathtt{=sin~x+xcos~x}$

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

$\displaystyle \mathtt{4)~f(x)= \frac{1}{x}}\\ \\ \mathtt{f'(x)=\left( \frac{1}{x} \right)'=\left(x^{-1}\right)'=(-1)\cdot x^{-1-1}=(-1)\cdot x^{-2}=(-1)\cdot \frac{1}{x^2}= - \frac{1}{x^2} }$

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

$\displaystyle\mathtt{6)~f(x)= \frac{1+e^x}{2+e^x} }\\ \\ \mathtt{f'(x)=\left( \frac{1+e^x}{2+e^x} \right)'= \frac{(1+e^x)'\cdot(2+e^x)-(1+e^x)\cdot(2+e^x)'}{(2+e^x)^2}=}\\\\ \mathtt{=\frac{(1'+(e^x)')\cdot(2+e^x)-(1+e^x)\cdot(2'+(e^x)')}{(2+e^x)'}=}\\\\\mathtt{= \frac{(0+e^x)\cdot(2+e^x)-(1+e^x)\cdot(0+e^x)}{(2+e^x)^2}=\frac{e^x(2+e^x)-(1+e^x)e^x}{(2+e^x)^2}=}\\\\\mathtt{=\frac{2e^x+e^{2x}-e^x-e^{2x}}{(2+e^x)^2}= \frac{e^x}{(2+e^x)^2}}$