Me ajudam so a resolver esse exercicio de Derivada f(x)=x/x+1

Derivada de produto de:

 f(x)=x/x+1

f'(x)=[(x)'(x+1)-x(x+1)']/(x+1)²

f'(x)=[(1)(x+1)-x(1)]/(x+1)²

f'(x)=[x+1-x]/(x+1)²

f'(x)=1/(x+1)²

Queremos calcular

\(\frac{d}{dx}\left(\frac{x}{x+1}\right)\)

Para tanto, vamos utilizar a Regra do Quociente, dada por:

\(\left(\frac{f}{g}\right)'=\frac{f'\cdot g-g'\cdot f}{g^2}\)

Assim:

\(f=x\\ g=x+1\)

\(\left(\frac{f}{g}\right)'=\frac{f'\cdot g-g'\cdot f}{g^2}\\ \left(\frac{x}{x+1}\right)'=\frac{1\cdot (x+1)-(1+0)\cdot.x}{(x+1)^2}\\ \left(\frac{x}{x+1}\right)'=\frac{(x+1-x}{(x+1)^2}\\ \left(\frac{x}{x+1}\right)'=\frac{1}{(x+1)^2}\\ \)

Assim

\(\boxed{\frac{d}{dx}\left(\frac{x}{x+1}\right)=\frac{1}{\left(x+1\right)^2}}\)