Question

Com o auxílio da tabela de derivadas imediatas, calcule a derivada primeira y’, da seguinte função: y =In(1 x/1-x).

Answer 1

Sendo u = 1+x/1-x

f(x) = ln(u)

f'(x) = ln(u) * u'

f'(x) = ln(1+x/1-x) * [1+x]'[1-x] - [1+x][1-x]' / [1-x]² 

f'(x) = ln(1+x/1-x) * [1][1-x] - [1+x][-1] / [1-x]² 

f'(x) = ln(1+x/1-x) * [1-x] + [1+x] / [1-x]² 

f'(x) = ln(1+x/1-x) * [1-x] + 1] / [1-x] 

Answer 2

pela tabela
$\boxed{\boxed{ln(u) ' = \frac{1}{u}*u' }}$

temos
$y=ln\left( \frac{1+x}{1-x} \right)$

logo 
$U = \frac{1+x}{1-x}$

para achar U' vc vai ter que derivar isso usando a regra do quociente
$( \frac{A}{B} )' = \frac{A'*B - A*B'}{B^2}$

temos
A = 1+x
A' = 0+1 = 1
B= 1-x
B' = 0-1 = -1

substituindo na regra do quociente
$U' = \frac{1*(1-x)-(1+x)*(-1)}{(1-x)^2} \\\\\\U' = \frac{1-x+1+x}{(1-x)^2} \\\\U'= \frac{2}{(1-x)^2}$


agora ja da pra calcular a derivada
$y'= \frac{1}{\left( \frac{1+x}{1-x} \right)} * \frac{2}{(1-x)^2} \\\\\\y'= \frac{(1-x)}{(1+x)}* \frac{2}{(1-x)^2} \\\\y'= \frac{1}{(1+x)}* \frac{2}{(1-x)} \\\\y'= \frac{2}{1^2-x^2} \\\\\boxed{\boxed{y'= \frac{2}{1-x^2} }}$