"One way to do this conveniently is by using properties of"
"logarithms."
"We want to compute:"
\qquad \qquad \qquad \qquad \qquad \qquad d/{dx} ln( e^x ( x - 1/x + 1)^{3/2} ).
"Let me take the function in question, by itself, first:"
\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ ln( e^x ( x - 1/x + 1)^{3/2} ).
"Using fundamental properties of logarithms, we can"
"simplify" \ \ f(x) \ \ "greatly:"
\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ ln( e^x ( x - 1/x + 1)^{3/2} )
\quad = \ ln( e^x ) + ln( x - 1/x + 1)^{3/2}; \qquad \ \ ln( A cdot B ) = ln(A) + ln(B)
\quad = \ x ln( e ) + 3/2 ln( x - 1/x + 1); \quad \ ln( A^r ) = r ln(A)
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ x ln( e ) + 3/2 ln( x - 1/x + 1)
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ x cdot 1 + 3/2 ln( x - 1/x + 1)
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = \ x + 3/2 ln( x - 1/x + 1).
"So:"
\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ x + 3/2 ln( x - 1/x + 1).
"Now, finding the derivative is easy [except for simplification !!]:"
\qquad \qquad \qquad \qquad \qquad \quad f'(x) \ = \ [ x ]' + [ 3/2 ln( x - 1/x + 1) ]'
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 1 + 3/2 [ ln( x - 1/x + 1) ]'
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 1 + 3/2 [ 1/( x - 1/x + 1)( x - 1/x + 1)' ]
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 1 + 3/2 [ 1/( x - 1/x + 1)( [ x ]' - [ 1/x ] + 0) ]
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 1 + 3/2 [ 1/( x - 1/x + 1)( 1 - [ x^{-1} ]' + 0) ]
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 1 + 3/2 [ 1/( x - 1/x + 1)( 1 - [ (-1) x^{-2} ] ) ]
\qquad \qquad "now simplify:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 1 + 3/2 [ 1/( x - 1/x + 1)( 1 - [ (-1) / x^{2} ] ) ]
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 1 + 3/2 [ 1/( x - 1/x + 1)( 1 + 1 / x^{2} ) ]
\qquad \qquad \qquad \qquad= \ 1 + 3/2 [ ( 1/( x - 1/x + 1 ) cdot x/x )( x^2 / x^2 + 1 / x^{2} ) ]
\qquad \qquad \qquad \qquad= \ 1 + 3/2 [ ( color{red}cancel{x}/( x^2 - 1 + x ) )( ( x^2 + 1 ) / x^{ color{red}cancel{2} } ) ]
\qquad \qquad \qquad \qquad= \ 1 + 3/2 [ ( 1/( x^2 - 1 + x ) )( ( x^2 + 1 ) / x ) ]
\qquad \qquad \qquad \qquad= \ 1 + 3/2 [ ( x^2 + 1)/( x (x^2 + x - 1 ) ) ]
\qquad \qquad \qquad \qquad= \ 1 + { 3 ( x^2 + 1) }/( 2 x (x^2 + x - 1 ) )
\qquad \qquad \qquad \qquad= \ ( 2 x (x^2 + x - 1 ) )/( 2 x (x^2 + x - 1 ) ) + { 3 ( x^2 + 1) }/( 2 x (x^2 + x - 1 ) )
\qquad \qquad \qquad \qquad= \ { 2 x (x^2 + x - 1 ) + 3 ( x^2 + 1) }/( 2 x (x^2 + x - 1 ) )
\qquad \qquad \qquad \qquad= \ { 2 x^3 + 2 x^2 - 2 x + 3 x^2 + 3 }/( 2 x (x^2 + x - 1 ) )
\qquad \qquad \qquad \qquad= \ { 2 x^3 + 5 x^2 - 2 x + 3 }/( 2 x (x^2 + x - 1 ) ).
"Sorry for all the work -- the calculus part was shorter,"
"the simplication long (not uncommon) !!"
"Summarizing:"
\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ ln( e^x ( x - 1/x + 1)^{3/2} ).
\qquad \qquad \qquad \qquad \qquad \quad f'(x) \ = \ { 2 x^3 + 5 x^2 - 2 x + 3 }/( 2 x (x^2 + x - 1 ) ).