Answer:
[tex]\frac{1}{(x+1)^2}[/tex]
Step-by-step explanation:
Differentiate using the Quotient rule.
Given
y = [tex]\frac{f(x)}{g(x)}[/tex] , then
[tex]\frac{dy}{dx}[/tex] = [tex]\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}[/tex]
Here
y = [tex]\frac{x}{x+1}[/tex] , with
f(x) = x ⇒ f'(x) = 1
g(x) = x + 1 ⇒ g'(x) = 1
Thus
[tex]\frac{dy}{dx}[/tex] = [tex]\frac{(x+1).1-x.1}{(x+1)^2}[/tex]
= [tex]\frac{x+1-x}{(x+1)^2}[/tex]
= [tex]\frac{1}{(x+1)^2}[/tex]
