Given $f ( x ) = \frac { x ^ { 3 } } { 2 x + 1 }$

Short Answer

Given $f ( x ) = \frac { x ^ { 3 } } { 2 x + 1 }$ , $g ( x ) = \frac { x ^ { 2 } + 2 } { 3 x ^ { 2 } }$ , and h(x)= f(x)g(x), find h'(1).Round to 2 decimal places.