Answer
$f’\left( x\right) =\dfrac {1}{\left( x+1\right) ^{2}}$
$f’\left( 3\right) =\dfrac {1}{16}$
$f’\left( 4\right) =\dfrac {1}{25}$
Work Step by Step
İn a function like $f(x)=\frac{A}{B}$ derivative of function is:
$f^{,}\left( x\right) =\dfrac {A'\times B-B'\times A}{B^{2}}$
$f\left( x\right) =\dfrac {x}{x+1}\Rightarrow f’\left( x\right) =\dfrac {\left( x\right)’ \times \left( x+1\right) -\left( x+1\right)’ \times x}{\left( x+1\right) ^{2}}=\dfrac {\left( 1\times x^{1-1}\right) \left( x+1\right) -\left( 1\times x^{1-1}+0\right) \times x}{\left( x+1\right) ^{2}}=\dfrac {1}{\left( x+1\right) ^{2}}$
$f’{}\left( 3\right) =\dfrac {1}{\left( x+1\right) ^{2}}=\dfrac {1}{\left( 3+1\right) ^{2}}=\dfrac {1}{4^{2}}=\dfrac {1}{16}$
$f’\left( 4\right) =\dfrac {1}{\left( x+1\right) ^{2}}=\dfrac {1}{\left( 4+1\right) ^{2}}=\dfrac {1}{5^{2}}=\dfrac {1}{25}$
