Question

Encontre as seguintes Derivadas :

Answer 1

$\boxed{\boxed{y= \sqrt{\frac{1+x}{1-x}} }}$

derivando
$\boxed{\boxed{\sqrt{u} = \frac{1}{2 \sqrt{u} } *u'}}$

$u = \frac{1+x}{1-x}$

para achar u'
vai ter que utilizar a regra da cadeia
$U'=( \frac{F}{G} )' = \frac{F'*G - F*G'}{G^2}$


$F = 1+x\\\\F'= 1\\\\G = 1-x\\\\G' =-1$

a derivada de u  fica
$\frac{1*(1-x) - (1+x)*(-1)}{(1-x)^2}= \frac{(1-x)+(1+x)}{(1-x)^2}= \frac{2}{(1-x)^2}$

substituindo na derivada da função 

$y'= \frac{1}{2 \sqrt{u} }*u'\\\\y'= \frac{1}{2 \sqrt{ \frac{1+x}{1-x} } }* \frac{2}{ (x-1)^2 } \\\\ \boxed{\boxed{y'= \frac{1}{(x-1)^2* \sqrt{ \frac{1+x}{1-x} }} }}$
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$y= \frac{2x^2-1}{x \sqrt{1+x^2} }$

usando a regra do quociente
$F = 2x^2-1\\\\F'=4x$

e
$G=x \sqrt{1+x^2}$

para derivar G...precisa usar a regra do produto
$(U*V) = U'*V + U*V'$

$U=x \\\ U'=1$

$V = \sqrt{1+x^2} \\\\V'= \frac{1}{2 \sqrt{1+x^2} } * 2x = \frac{x}{ \sqrt{1+x^2} }$

colocando na regra do produto
$1* \sqrt{1+x^2} + x* \frac{x}{ \sqrt{1+x^2} } \\\\=\sqrt{1+x^2} + \frac{x^2}{\sqrt{1+x^2} } \\\\= \frac{(\sqrt{1+x^2})^2+x^2}{\sqrt{1+x^2}} \\\\= \frac{1+x^2+x^2}{\sqrt{1+x^2}}\\\\ \boxed{G'= \frac{2x^2+1}{ \sqrt{1+x^2} } }$

voltando la pra regra do quociente
e substituindo os valores de f;f' ;g ;g'
$y' = \frac{4x *(x \sqrt{1+x^2} )- (2x^2-1)* (\frac{2x^2+1}{\sqrt{1+x^2}})}{(x \sqrt{1+x^2} )^2}$

resolvendo o numerador
fica
$4x^2 \sqrt{1+x^2}- \frac{(2x^2+1)*(2x^2-1)}{ \sqrt{1+x^2} }\\\\ =\frac{(4x^2)* (\sqrt{1+x^2}) * (\sqrt{1+x^2}) -(2x^2+1)*(2x^2-1)}{ \sqrt{1+x^2} } \\\\=\frac{(4x^2)*(1+x^2) -(4x^4-1)}{ \sqrt{1+x^2} }\\\\= \frac{(4x^2+4x^4)-(4x^4-1)}{ \sqrt{1+x^2} } \\\\= \frac{4x^2+1}{ \sqrt{1+x^2} }$

resolvendo o denominador
$(x \sqrt{1+x^2} )^2 =x^2*( \sqrt{1+x^2})^2$

a derivada fica

$y'= \frac{ \frac{4x^2+1}{ \sqrt{1+x^2} }}{x^2*( \sqrt{1+x^2})^2 } \\\\y'= \frac{4x^2+1}{ (\sqrt{1+x^2} )*x^2*( \sqrt{1+x^2})^2 } \\\\\\\boxed{\boxed{y'= \frac{4x^2+1}{x^2( \sqrt{x+1})^3 } }}$