Question

Ma poate ajuta cineva cu cateva exercitii?

Answer 1

$1.~~~a)~f(x)=x^2-5x+1\\ \\f'(x)=(x^2-5x+1)'=(x^2)'-(5x)'+1'=2x-5 \cdot x'+0=\\ \\=2x-5 \cdot 1=2x-5\\ \\ b)~f(x)=-3x^2+2x^4-0,5\\ \\f'(x)=(-3x^2+2x^4-0,5)'=(-3x^2)'+(2x^4)'-(0,5)'=\\ \\=-3\cdot (x^2)'+2 \cdot (x^4)'-0=-3 \cdot 2x+2\cdot4x^3=-6x+8x^3\\ \\ c)~f(x)=3x-x^6\\ \\ f'(x)=(3x-x^6)'=(3x)'-(x^6)'=3 \cdot x'-6x^5=3 \cdot 1-6x^5=3-6x^5$

$\displaystyle d)~f(x)=2e^x-ln~x+ \sqrt{x} ,~x\in(0,+\infty)\\\\ f'(x)=(2e^x-ln~x+ \sqrt{x} )'=(2e^x)'-(ln~x)'+( \sqrt{x} )'=\\\\ =2 \cdot (e^x)'- \frac{1}{x} + \frac{1}{2 \sqrt{x} } =2e^x- \frac{1}{x} + \frac{1}{2 \sqrt{x} }\\ \\ e)~f(x)=2^x+3^x- \sqrt[3]{x},~x\in(0,+\infty)\\ \\ f'(x)=(2^x+3^x- \sqrt[3]{x} )'=(2^x)'+(3^x)'-( \sqrt[3]{x} )'=\\ \\ =2^xln2+3^xln3- \frac{1}{3 \sqrt[3]{x^2} }$

$\displaystyle f)~f(x)=x^5-sin~x+2cos~x\\ \\ f'(x)=(x^5-sin~x+2cos~x)'=(x^5)'-(sin~x)'+(2cos~x)'=\\ \\ =5x^4-cos~x+2 \cdot (cos~x)'=5x^4-cos~x-2sin~x \\ \\ g)~f(x)=(x-3)^2+4=x^2-2 \cdot x \cdot 3+3^2+4=x^2-6x+13\\ \\ f'(x)=(x^2-6x+13)'=(x^2)'-(6x)'+(13)'=2x-6 \cdot x'+0=\\ \\ =2x-6\cdot 1=2x-6$

$\displaystyle h)~f(x)=(x+1)^2+(x-2)^3\\ \\ f'(x)=[(x+1)^2+(x-2)^3]'=[(x+1)^2]'+[(x-2)^3]'=\\ \\ =2(x+1)(x+1)'+3(x-2)^2(x-2)'=\\ \\ =2(x+1) \cdot 1+3(x^2-2 \cdot x \cdot 2+2^2) \cdot 1=2(x+1)+3(x^2-4x+4)=\\ \\ =2x+2+3x^2-12x+12=3x^2-10x+14$

$\displaystyle i)~f(x)=2 \sqrt{x} +x \sqrt{2} ,~x\in(0,+\infty)\\ \\ f'(x)=(2 \sqrt{x} +x \sqrt{2} )'=(2 \sqrt{x} )'+(x \sqrt{2} )'=2 \cdot ( \sqrt{x} )'+ \sqrt{2} \cdot x'=\\ \\ =\not2 \cdot \frac{1}{\not2 \sqrt{x} }+ \sqrt{2} \cdot 1= \frac{1}{ \sqrt{x} } + \sqrt{2} \\ \\ j)~f(x)=log_3x+2^x-x,~x\in(0,+\infty)\\ \\ f'(x)=(log_3x+2^x-x)'=(log_3x)'+(2^x)'-x'=\frac{1}{xln3} +2^xln2-1$

$\displaystyle k)~f(x)=- \sqrt{x} -3ln~x+3^x,~x\in(0,+\infty)\\ \\ f'(x)=(- \sqrt{x} -3ln~x+3^x)'=(- \sqrt{x} )'-(3ln~x)'+(3^x)'=\\ \\ =-( \sqrt{x} )'-3 \cdot (ln~x)'+3^xln3=- \frac{1}{2 \sqrt{x} } -3 \cdot \frac{1}{x} +3^xln3=\\ \\ =- \frac{1}{2 \sqrt{x} } - \frac{3}{x} +3^xln3$

$\displaystyle l)~f(x)= \sqrt{5} -5 \sqrt{x} +5cos~x,~x\in(0,+\infty)\\ \\ f'(x)=( \sqrt{5} -5 \sqrt{x} +5cos~x)'=( \sqrt{5} )'-(5 \sqrt{x} )'+(5cos~x)'=\\ \\ =0-5 \cdot ( \sqrt{x} )'+5 \cdot (cos~x)'=-5 \cdot \frac{1}{2 \sqrt{x} } +5\cdot (-sin~x)=\\ \\ =- \frac{5}{2 \sqrt{x} } -5sin~x$

$\displaystyle m)~f(x)=10^x+7ln~x-3sin~x,~x\in(0,+\infty)\\ \\ f'(x)=(10^x+7ln~x-3sin~x)'=(10^x)'+(7ln~x)'-(3sin~x)'=\\ \\ =10^xln10+7\cdot(ln~x)'-3 \cdot (sin~x)'=10^xln10+7 \cdot \frac{1}{x} -3\cdot cos~x=\\ \\ =10^xln10+ \frac{7}{x} -3cos~x$

$\displaystyle n)~f(x)=2011^x-xln2011\\ \\ f'(x)=(2011^x-xln2011)'=(2011^x)'-(xln2011)'=\\ \\ =2011^xln2011-ln2011\cdot 1=2011^xln2011-ln2011\\ \\ p)~f(x)=ln~4+lg~x-e^{x+1},~x\in(0,+\infty)\\ \\ f'(x)=(ln~4+lg~x-e^{x+1})'=(ln~4)'+(lg~x)'-(e^{x+1})'=\\ \\ =0+ \frac{1}{xln10} -e^{x+1}\cdot (x+1)'= \frac{1}{xln10} -e^{x+1} \cdot (x'+1')=\\ \\ = \frac{1}{xln10} -e^{x+1}\cdot 1= \frac{1}{xln10} -e^{x+1}$

$\displaystyle q)~f(x)=f(x)=log_2x^3-log_3x^2,~x\in(0,+\infty)\\ \\ f'(x)=(log_2x^3-log_3x^2)'=(log_2x^3)'-(log_3x^2)'= \frac{3}{xln2} - \frac{2}{xln3} \\ \\ r)~f(x)=2^{2x}+3^{3x}\\ \\ f'(x)=(2^{2x}+3^{3x})'=(2^{2x})'+(3^{3x})=\\ \\ =2^{2x}\cdot (2x)'\cdot ln2+3^{3x}\cdot (3x)'+ln3=2^{2x}\cdot 2x'\cdot ln2+3^{3x}\cdot 3x'+ln3=\\ \\ =2^{2x}\cdot 2 \cdot 1 \cdot ln2+3^{3x}\cdot 3\cdot 1+ln~3=2^{2x}\cdot 2ln2+3^{3x}\cdot 3ln3=\\ \\ =2^{2x+1}ln2+3^{3x+1}ln3$

$\displaystyle 2.~~~a)~f(x)=xe^x\\ \\ f'(x)=(xe^x)'=x'e^x+x(e^x)'=1 \cdot e^x+x \cdot e^x=e^x+xe^x\\ \\ b)~f(x)=(x-2)ln~x\\ \\ f'(x)=[(x-2)ln~x]'=(x-2)'lnx+(x-2)(ln~x)'=\\\\=1 \cdot ln~x+(x-2) \cdot \frac{1}{x} =ln~x+ \frac{x-2}{x} \\ \\ c)~f(x)=x^2\cdot2^x\\ \\ f'(x)=(x^2\cdot 2^x)'=(x^2)'\cdot2^x+x^2\cdot (2^x)'=2x \cdot 2^x+x^2 \cdot 2^xln2=\\ \\ =2^{1+x}x+x^2 \cdot 2^xln2$

$\displaystyle d)~f(x)=(3x-1) \sqrt{x} \\ \\ f'(x)=[(3x-1) \sqrt{x} ]'=(3x-1)' \cdot \sqrt{x} +(3x-1)\cdot ( \sqrt{x} )'=\\ \\ =3 \sqrt{x}+(3x-1) \cdot\frac{1}{2\sqrt{x} }=3 \sqrt{x} + \frac{3x-1}{2 \sqrt{x}}\\\\e)~f(x)= \sqrt[3]{x}(1+\sqrt{x} )\\\\f'(x)=[ \sqrt[3]{x} (1+ \sqrt{x} )]'=( \sqrt[3]{x} )'\cdot(1+ \sqrt{x} )+\sqrt[3]{x}(1+ \sqrt{x} )'=\\ \\ = \frac{1}{3 \sqrt[3]{x^2}}\cdot(1+\sqrt{x} )+ \sqrt[3]{x}\cdot \frac{1}{2 \sqrt{x} } = \frac{1+ \sqrt{x} }{3 \sqrt[3]{x^2}}+\frac{\sqrt[3]{x}}{2\sqrt{x}}$

$\displaystyle f)~f(x)=(2x^3-x+3)\cdot 3^x\\ \\ f'(x)=[(2x^3-x+3)\cdot 3^x]'=(2x^3-x+3)'\cdot 3^x+(2x^3-x+3)(3^x)'=\\ \\ =(3 \cdot 2x^2-1) \cdot 3^x+(2x^3-x+3) \cdot 3^xln3=\\ \\ =(6x^2-1) \cdot 3^x+(2x^3-x+3) \cdot 3^xln3$

$\displaystyle g)~f(x)=2xsin~x\\ \\ f'(x)=(2xsin~x)'=2(xsin~x)'=2(x' \cdot sin~x+x \cdot (sin~x)')=\\ \\ =2(1 \cdot sin~x+x \cdot cos~x)=2(sin~x+xcos~x)=2sin~x+2xcos~x \\ \\ h)~f(x)=x^4log_2x\\ \\ f'(x)=(x^4log_2x)'=(x^4)'\cdot log_2x+x^4 \cdot (log_2x)=\\ \\ =4x^3 \cdot log_2x+x^4 \cdot \frac{1}{xln2} =4x^3log_2x+ \frac{x^4}{xln2}$