Question

1- determine as derivadas das funções abaixo:
a- x sen x

b- g(x)= (x+3) tgx

c- h(x)= raiz de x²-1

d- f(x)= In(x³) cos (2x)

3- g(x)= x²
e
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sen 2x

4- f(x)= tgx cos x

Answer 1

$y=x*sen(x)$

utilizando a regra do produto 
u = x
u' = 1
v = sen(x) 
v' =cos(x)

$y' = U'*V + U*V'\\\\y'= 1*sen(x)+x*cos(x)\\\\y'=sen(x)+x*cos(x)$

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b) $g(x) = (x+3)*tg(x)$
mesma coisa
u = x+3
u' = 1 +0 =1
v = tg(x)
v' = sec²(x)
$g'(x)= 1*tg(x) + (x+3)*sec^2(x)\\\\g'(x)=tg(x)+(x+3)*sec^2(x)$

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c) deriva de $\sqrt{u} = \frac{1}{2 \sqrt{u} }*u'$
aplicando isso
$h(x)= \sqrt{x^2-1} \\\\h'(x)= \frac{1}{2 \sqrt{x^2-1} }*(x^2-1)'\\\\h'(x)= \frac{1}{2 \sqrt{x^2-1} } *(2x-0)\\\\h'(x)= \frac{2x}{2 \sqrt{x^2-1} } = \frac{x}{\sqrt{x^2-1}}$

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d)
$f(x) = ln(x^3)*cos(2x)$

aplicando a regra do produto
$U = ln(x^3)\\\\U' = \frac{1}{x^3}*(x^3)' \\\\U'= \frac{1}{x^3}*(3x^2)\\\\U' = \frac{3}{x}$

$V=cos(2x)\\\\V'=-sen(2x)*(2x)'\\\\V'=-sen(2x)*(2*1)\\\\V'=-2sen(2x)$

colocando na regra do produto
$f'(x) = \frac{3}{x}* cos(2x) + ln(x^3)*(-2sen(2x))\\\\f'(x)= \frac{3cos(2x)}{x}-sen(2x)*ln(x^3)$

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e) acho que seria e^(x²)/sen(2x))

$g(x) = \frac{e^{(x^2)}}{sen(2x)}$

regra do quociente
$( \frac{U}{V} )' = \frac{U'*V - U*V'}{V^2}$

temos
$U=e^{x^2}\\\\U'=e^{x^2}*(x^2)'\\\\U'=e^{x^2}*2x\\\\U' = 2xe^{x^2}$

$V=sen(2x)\\\\V'=cos(2x)*(2x')\\\\V'=2cos(2x)$

colocando na regra do quociente
$g'(x)= \frac{2xe^{x^2}*sen(2x) - e^{x^2}*2cos(2x)}{(sen(2x))^2} \\\\g'(x)=\frac{2e^{x^2}*[x*sen(2x) - cos(2x)]}{(sen(2x))^2}$
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f) usando a regra do produto
 $f(x)=tg(x)*cos(x)\\\\f'(x)=sec^2(x)*cos(x) -sen(x)*tg(x)$