# "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 ) ). #
