Question

am si eu nevoie de ajutor cu cateva functii va rog dau coroana primului care imi raspunde

Answer 1

$\displaystyle \mathtt{a)~f(x)=\left(x^2+1\right)^5}\\ \\ \mathtt{f'(x)=\left[\left(x^2+1\right)^5\right]'=5\cdot\left(x^2+1\right)^{4} \cdot \left(x^2+1\right)'=5\left(x^2+1\right)^4\cdot(2x+0)=}\\ \\ \mathtt{=5\left(x^2+1\right)^4 \cdot 2x=10x\left(x^2+1\right)^4}$

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

$\displaystyle \mathtt{c)~f(x)=e^{3x}}\\ \\ \mathtt{f'(x)=\left(e^{3x}\right)'=e^{3x}\cdot (3x)'=e^{3x}\cdot 3=3e^{3x}}$

$\displaystyle \mathtt{d)~f(x)=sin5x}\\ \\ \mathtt{f'(x)=(sin5x)'=cos5x\cdot(5x)'=cos5x\cdot5=5cos5x}$

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

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

$\displaystyle \mathtt{g)~f(x)= \sqrt[\mathtt{3}]{\mathtt{x+sin~x}}}\\ \\ \mathtt{f'(x)= \left( \sqrt[\mathtt3]{\mathtt{x+sin~x}}\right)'= \frac{1}{3(x+sin~x)^{ \frac{2}{3}}} \cdot (x+sin~x)'=}\\ \\ \mathtt{= \frac{(x+sin~x)'}{3(x+sin~x)^{ \frac{2}{3} }}= \frac{x'+(sin~x)'}{3(x+sin~x)^{ \frac{2}{3} }}= \frac{1+cos~x}{3(x+sin~x)^{ \frac{2}{3} }} }$

$\displaystyle \mathtt{h)~f(x)=tg^2x+cos2x}\\ \\ \mathtt{f'(x)=\left(tg^2x+cos2x\right)'=\left(tg^2x\right)'+(cos2x)'=}\\ \\ \mathtt{=(tg~x\cdot tg~x)'+2(cos~x)'=(tg~x)' \cdot tg~x+tg~x \cdot (tg~x)'-2sin2x=}\\ \\ \mathtt{=sec^2x \cdot tg~x+tg~x \cdot sec^2x-2sin2x=2sec^2x\cdot tg~x-2sin2x}$

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

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

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

$\displaystyle \mathtt{l)~f(x)=(2sin~x-3)^6}\\ \\ \mathtt{f'(x)=\left[(2sin~x-3)^6\right]'=6\cdot(2sin~x-3)^5 \cdot (2sin~x-3)'=}\\ \\ \mathtt{=6(2sin~x-3)^5 \cdot 2(sin~x)'-3'=6(2sin~x-3)^5 \cdot 2cos~x=}\\ \\ \mathtt{=12cos~x(2sin~x-3)^5}$

$\displaystyle \mathtt{m)~f(x)=\left( \frac{2}{x+1}\right)^5 }\\ \\ \mathtt{f'(x)=\left[\left( \frac{2}{x+1}\right)^5\right]'=5 \cdot \left( \frac{2}{x+1} \right)^4 \cdot \left( \frac{2}{x+1}\right)'= }\\ \\ \mathtt{=5 \cdot \left( \frac{2}{x+1}\right)^4 \cdot \frac{2' \cdot (x+1)-2 \cdot (x+1)'}{(x+1)^2}=5 \cdot \frac{16}{(x+1)^4} \cdot \frac{-2}{(x+1)^2}= }\\ \\ \mathtt{=- \frac{160}{(x+1)^6} }$

$\displaystyle \mathtt{n)~f(x)=sin^32x}\\ \\ \mathtt{f'(x)=\left(sin^32x\right)'=3sin^22x \cdot (sin2x)'=6sin^22x \cdot cos2x}$