Question

algm me ajuda? ficarei muito grato pela resposta
Utilize as regras de derivação para determinar:

Answer 1

a)

$f'(x)=\frac{(x+1)(cosx)-(senx)(1)}{(x+1)^2} =\frac{(x+1)cos(x)-sen(x)}{(x+1)^2}$

$f'(0)=\frac{(0+1)cos(0)-sen(0)}{(0+1)^2} =\frac{1-0}{1} =1$

b)

$f'(x)=(6x)(e^x)+(3x^2+1)e^x$

$f'(2)=(6.2)(e^2)+e^2(3(2)^2+1)=12e^2+13e^2=25e^2=184,72$

c)

$f(x)=1+\sqrt{x} \\f'(x)=x^\frac{1}{2} +1=\frac{x^\frac{-1}{2} }{2} +0=\frac{1}{2x^\frac{1}{2} } =\frac{1}{2\sqrt{x} }$

$f'(3)=\frac{1}{2\sqrt{3} } =0,28$

d)

$f'(x)=(1.cotx+(-cossec^2(x))x)-(2(1.secx+x.secxtgx)\\f'(x)=cotx-xcossec^2x+2secx(1+x.tgx)$

Answer 2

Explicação passo-a-passo:

a)

$\sf f(x)=\dfrac{sen~x}{x+1}$

Regra do quociente:

$\sf \Big(\dfrac{f}{g}\Big)'=\dfrac{f'\cdot g-f\cdot g'}{g^2}$

$\sf f'(x)=\dfrac{(sen~x)'\cdot(x+1)-sen~x\cdot(x+1)'}{(x+1)^2}$

$\sf f'(x)=\dfrac{cos~x\cdot(x+1)-sen~x\cdot1}{(x+1)^2}$

$\sf f'(x)=\dfrac{(x+1)\cdot cos~x-sen~x}{(x+1)^2}$

Assim:

$\sf f'(0)=\dfrac{(0+1)\cdot cos~0-sen~0}{(0+1)^2}$

$\sf f'(0)=\dfrac{1\cdot1-0}{1^2}$

$\sf f'(0)=\dfrac{1-0}{1}$

$\sf f'(0)=\dfrac{1}{1}$

$\sf \red{f'(0)=1}$

b)

$\sf f(x)=(3x^2+1)\cdot e^x$

Regra do produto:

$\sf (f\cdot g)'=f'\cdot g+f\cdot g'$

$\sf f'(x)=(3x^2+1)'\cdot e^x+(3x^2+1)\cdot(e^x)'$

$\sf f'(x)=2\cdot3x\cdot e^x+(3x^2+1)\cdot e^x$

$\sf f'(x)=6x\cdot e^x+(3x^2+1)\cdot e^x$

$\sf f'(x)=e^x\cdot[6x+(3x^2+1)]$

$\sf f'(x)=e^x\cdot(3x^2+6x+1)$

Assim:

$\sf f'(2)=e^2\cdot(3\cdot2^2+6\cdot2+1)$

$\sf f'(2)=e^2\cdot(3\cdot4+6\cdot2+1)$

$\sf f'(2)=e^2\cdot(12+12+1)$

$\sf \red{f'(2)=25e^2}$

c)

$\sf f(x)=\dfrac{\sqrt[3]{x}+x}{\sqrt{x}}$

Regra do quociente:

$\sf \Big(\dfrac{f}{g}\Big)'=\dfrac{f'\cdot g-f\cdot g'}{g^2}$

$\sf f'(x)=\dfrac{(\sqrt[3]{x}+x)'\cdot\sqrt{x}-(\sqrt[3]{x}+x)\cdot(\sqrt{x})'}{(\sqrt{x})^2}$

$\sf f'(x)=\dfrac{(x^{\frac{1}{3}}+x)'\cdot\sqrt{x}-(\sqrt[3]{x}+x)\cdot(x^{\frac{1}{2}})'}{x}$

$\sf f'(x)=\dfrac{\Big(\frac{1}{3}\cdot x^{\frac{1}{3}-1}+1\Big)\cdot\sqrt{x}-(\sqrt[3]{x}+x)\cdot\Big(\frac{1}{2}\cdot x^{\frac{1}{2}-1}\Big)}{x}$

$\sf f'(x)=\dfrac{\Big(\frac{1}{3}\cdot x^{\frac{-2}{3}}+1\Big)\cdot\sqrt{x}-(\sqrt[3]{x}+x)\cdot\Big(\frac{1}{2}\cdot x^{\frac{-1}{2}}\Big)}{x}$

$\sf f'(x)=\dfrac{\Big(\frac{1}{3}\cdot\frac{1}{\sqrt[3]{x^2}}+1\Big)\cdot\sqrt{x}-(\sqrt[3]{x}+x)\cdot\Big(\dfrac{1}{2}\cdot\frac{1}{\sqrt{x}}\Big)}{x}$

$\sf f'(x)=\dfrac{\Big(\frac{1}{3\sqrt[3]{x^2}}+1\Big)\cdot\sqrt{x}-\Big[(\sqrt[3]{x}+x)\cdot\frac{1}{2\sqrt{x}}\Big]}{x}$

Assim:

$\sf f'(3)=\dfrac{\Big(\frac{1}{3\sqrt[3]{3^2}}+1\Big)\cdot\sqrt{3}-\Big[(\sqrt[3]{3}+3)\cdot\frac{1}{2\sqrt{3}}\Big]}{3}$

$\sf f'(3)=\dfrac{\frac{\sqrt{3}}{3\sqrt[3]{9}}+\sqrt{3}-\frac{\sqrt[3]{3}}{2\sqrt{3}}-\frac{3}{2\sqrt{3}}}{3}$

$\sf f'(3)=\dfrac{\frac{\sqrt{3}}{3\sqrt[3]{9}}\cdot\frac{\sqrt[3]{3}}{\sqrt[3]{3}}+\sqrt{3}-\frac{\sqrt[3]{3}}{2\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}-\frac{3}{2\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}}{3}$

$\sf f'(3)=\dfrac{\frac{\sqrt{3}\cdot\sqrt[3]{3}}{3\cdot3}+\sqrt{3}-\frac{\sqrt{3}\cdot\sqrt[3]{3}}{2\cdot3}-\frac{3\sqrt{3}}{2\cdot3}}{3}$

$\sf f'(3)=\dfrac{\frac{\sqrt{3}\cdot\sqrt[3]{3}}{9}+\sqrt{3}-\frac{\sqrt{3}\cdot\sqrt[3]{3}}{6}-\frac{3\sqrt{3}}{6}}{3}$

$\sf f'(3)=\dfrac{\frac{2\sqrt{3}\cdot\sqrt[3]{3}+18\sqrt{3}-3\sqrt{3}\cdot\sqrt[3]{3}-9\sqrt{3}}{18}}{3}$

$\sf f'(3)=\dfrac{\frac{\sqrt{3}\cdot(18-9+2\sqrt[3]{3}-3\sqrt[3]{3})}{18}}{3}$

$\sf f'(3)=\dfrac{\frac{\sqrt{3}\cdot(9-\sqrt[3]{3})}{18}}{3}$

$\sf f'(3)=\dfrac{\sqrt{3}\cdot(9-\sqrt[3]{3})}{18}\cdot\dfrac{1}{3}$

$\sf \red{f'(3)=\dfrac{\sqrt{3}\cdot(9-\sqrt[3]{3})}{54}}$

d)

$\sf f(x)=x\cdot cotg~x+2x\cdot sec~x$

Regra do produto:

$\sf (f\cdot g)'=f'\cdot g+f\cdot g'$

$\sf f'(x)=(x)'\cdot cotg~x+x\cdot(cotg~x)'+(2x)'\cdot sec~x+2x\cdot(sec~x)$

$\sf f'(x)=1\cdot cotg~x+x\cdot(-cossec^2~x)+2\cdot sec~x+2x\cdot sec~x\cdot tg~x$

$\sf \red{f'(x)=cotg~x-x\cdot cossec^2~x+2\cdot sec~x(1+x\cdot tg~x)}$