Question

Urgentt va roggg dau coronitee

Answer 1

$\displaystyle \mathtt{a)~f(x)=4x^3-2x^2+5}\\ \\ \mathtt{f'(x)=\left(4x^3-2x^2+5\right)'=\left(4x^3\right)'-\left(2x^2\right)'+5'=}\\ \\ \mathtt{4 \cdot \left(x^3\right)'-2 \cdot\left(x^2\right)'+0=4 \cdot 3x^2-2 \cdot 2x=12x^2-4x}$

$\displaystyle \mathtt{b)~f(x)=sin~x(sin~x-3)}\\ \\ \mathtt{f'(x)=[sin~x(sin~x-3)]'=}\\ \\ \mathtt{=(sin~x)'\cdot(sin~x-3)+sin~x \cdot (sin~x-3)'=}\\ \\ \mathtt{=cos~x \cdot(sin~x-3)+sin~x \cdot cos~x}$

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

$\displaystyle \mathtt{d)~f(x)=(1-2ln~x)(ln~x+2)}\\ \\ \mathtt{f'(x)=[(1-2ln~x)(ln~x+2)]'=}\\ \\ \mathtt{=(1-2ln~x)' \cdot (ln~x+2)+(1-2ln~x)\cdot(ln~x+2)'=}\\ \\ \mathtt{=1'-2(ln~x)' \cdot(ln~x+2)+(1-2ln~x) \cdot (ln~x)'+2'=}\\ \\ \mathtt{=-2 \cdot \frac{1}{x} \cdot (ln~x+2)+(1-2ln~x)\cdot \frac{1}{x}=}\\ \\ \mathtt{= \frac{-2ln~x-4+1-2ln~x}{x} = \frac{-4ln~x-3}{x} }$

$\displaystyle \mathtt{e)~f(x)=(sin~x+x)^2-2x}\\ \\ \mathtt{f'(x)=\left[(sin~x+x)^2-2x\right]'=\left[(sin~x+x)^2\right]'-(2x)'=}\\ \\ \mathtt{=2(sin~x+x)\cdot (sin~x+x)'-2=2(sin~x+x)(cos~x+1)-2}$

$\displaystyle \mathtt{f)~f(x)=2\cdot x\cdot ln~x\cdot sin~x}\\ \\ \mathtt{f'(x)=(2\cdot x\cdot ln~x\cdot sin~x)'=2(x'\cdot ln~x \cdot sin~x+x \cdot (ln~x \cdot sin~x)')=}\\ \\ \mathtt{=2\left(1 \cdot ln~x \cdot sin~x+x \cdot \left( \frac{1}{x}\cdot sin~x+ln~x \cdot cos~x\right)\right) =}\\ \\ \mathtt{=2(ln~x \cdot sin~x+sin~x+xln~x \cdot cos~x)=}\\ \\ \mathtt{=2ln~x \cdot sin~x+2sin~x+2xln~x \cdot cos~x}$

$\displaystyle \mathtt{g)~f(x)=\left(x^2+2\right)\cdot ln(x+1)}\\ \\ \mathtt{f'(x)=\left(x^2+2\right)'\cdot ln(x+1)+\left(x^2+2\right)\cdot[ln(x+1)]'=}\\ \\ \mathtt{=\left\left[\left(x^2\right)'+2'\right] \cdot ln(x+1)+\left(x^2+2\right) \cdot \frac{1}{x+1} = 2xln(x+1)+ \frac{x^2+2}{x+1} }$

$\displaystyle \mathtt{h)~f(x)=\left(x^2+2x\right) \cdot tg~x}\\ \\ \mathtt{f'(x)=\left[\left(x^2+2x\right)\cdot tg~x\right]'=\left(x^2+2x\right)' \cdot tg~x+(x^2+2x) \cdot (tg~x)'=}\\ \\ \mathtt{=\left[\left(x^2\right)'+(2x)'\right]\cdot tg~x+\left(x^2+2x\right) \cdot sec^2x=}\\ \\ \mathtt{=(2x+2) \cdot tg~x+\left(x^2+2x\right) \cdot sec^2x}$

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

$\displaystyle\mathtt{j)~ f(x)=\frac{x^2-x+2}{x^2-4x+3} }\\ \\ \mathtt{f'(x)=\left( \frac{x^2-x+2}{x^2-4x+3}\right) '=}\\ \\ \mathtt{= \frac{\left(x^2-x+2\right)' \cdot\left(x^2-4x+3\right)-\left(x^2-x+2\right)\cdot\left(x^2-4x+3\right)'}{\left(x^2-4x+3\right)^2}= }\\ \\ \mathtt{=\frac{(2x-1)\left(x^2-4x+3\right)-\left(x^2-x+2)(2x-4)}{\left(x^2-4x+3\right)^2}=}\\\\\mathtt{=\frac{2x^3-8x^2+6x-x^2+4x-3-2x^3+2x^2-4x+4x^2-4x+8}{\left(x^2-4x+3\right)^2}=}\\ \\\mathtt{= \frac{-3x^2+2x+5}{\left(x^2-4x+3\right)^2}}$

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

$\displaystyle \mathtt{l)~f(x)= \frac{xsin~x}{2+cos~x} }\\ \\ \mathtt{f'(x)=\left( \frac{xsin~x}{2+cos~x}\right)'= \frac{(xsin~x)'\cdot(2+cos~x)-xsin~x \cdot (2+cos~x)'}{(2+cos~x)^2} =}\\ \\ \mathtt{= \frac{\left(x' \cdot sin~x+x \cdot (sin~x)'\right)\cdot(2+cos~x)-xsin~x\cdot(-sin~x)}{(2+cos~x)^2}= }\\ \\ \mathtt{= \frac{(sin~x+xcos~x)(2+cos~x)+xsin^2x}{(2+cos~x)^2} }$

$\displaystyle \mathtt{m)~ f(x)=\frac{sin~x+cos~x}{sin~x-cos~x} }\\ \\ \mathtt{f'(x)=\left( \frac{sin~x+cos~x}{sin~x-cos~x}\right)'= }\\ \\ \mathtt{= \frac{(sin~x+cos~x)' (sin~x-cos~x)-(sin~x+cos~x)(sin~x-cos~x)'}{(sin~x-cos~x)^2} =}\\ \\ \mathtt{= \frac{(cos~x-sin~x)(sin~x-cos~x)-(sin~x+cos~x)(cos~x+sin~x)}{(sin~x-cos~x)^2}= }\\ \\ \mathtt{= \frac{2sin~x \cdot cos~x-sin^2x-cos^2x-1-sin2x}{sin^2x-2sin~x \cdot cos~x+cos^2x}= }$

$\displaystyle \mathtt{= \frac{2sin~x \cdot cos~x-1+cos^2x-cos^2x-1-2sin~x \cdot cos~x}{1-cos^2x-2sin~x \cdot cos~x+cos^2x} =} \\ \\ \mathtt{= \frac{-2}{1-2sin~x \cdot cos~x}=- \frac{2}{1-sin2x} }$