Question

f(x) derivat din x-1/x+1+ln(x+1)-lnx

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$f(x)=\frac{x-1}{x+1} +ln(x+1)-ln x\\f'(x)=(\frac{x-1}{x+1} )' + [ln(x+1)]' +(lnx)'=\\\frac{(x-1)' *(x+1)-(x+1)'*(x-1)}{(x+1)^{2} } +\frac{1}{x+1} * (x+1)'-\frac{1}{x} *x'=\\\frac{x+1-x+1}{(x+1)^{2} } +\frac{1}{x+1} -\frac{1}{x} =\\\frac{2}{(x+1)^{2} } +\frac{1}{x+1} +\frac{1}{x} =\\ \frac{2x+x(x+1)-(x+1)^{2}}{x((x+1)^{2}} =\\\frac{2x+x^{2}+x-x^{2}-2x-1}{x(x+1)^{2}} =\\\frac{x-1}{x(x+1)^{2}}$