Given:
[tex]f(x)=\frac{x}{x+1}[/tex]
To find: The first derivative
Explanation:
Using the quotient rule,
[tex](\frac{u}{v})^{\prime}=\frac{vu^{\prime}-uv^{\prime}}{v^2}[/tex]
Here,
[tex]\begin{gathered} u=x \\ u^{\prime}=\frac{d}{dx}(x) \\ \Rightarrow u^{\prime}=1 \\ v=x+1 \\ v^{\prime}=\frac{d}{dx}(x+1) \\ \Rightarrow v^{\prime}=1 \end{gathered}[/tex]
On substitution we get,
[tex]\begin{gathered} f^{\prime}(x)=\frac{(x+1)(1)-x(1)}{(x+1)^2} \\ f^{\prime}(x)=\frac{x+1-x}{(x+1)^2} \\ f^{\prime}(x)=\frac{1}{(x+1)^2} \end{gathered}[/tex]
Final answer: The first derivative of the given function is,
[tex]f^{\prime}(x)=\frac{1}{(x+1)^{2}}[/tex]
