How do you differentiate #g(x) = (1/x^3)*sqrt(1-e^(2x))# using the product rule?

1 Answer
Feb 13, 2018

Answer:

# \ #

# \qquad \qquad \qquad \qquad \qquad g'(x) \ = \ [ 3 + ( x- 3 ) e^{ 2x } ] / { x^{ 4 } \sqrt{ 1 - e^{ 2x } } } \quad. #

Explanation:

# \ #

# "We are given:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad g(x) \ = \ ( 1 / x^3 ) \cdot \sqrt{ 1 - e^{ 2x } }. #

# "We can rewrite" \ g(x) \ "to prepare it for differentiation:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad g(x) \ = \ x^{ -3 } \cdot ( 1 - e^{ 2x } )^{ 1/2 }. #

# "Using the Product Rule:" #

# \qquad \qquad g'(x) \ = \ x^{ -3 } \cdot [ ( 1 - e^{ 2x } )^{ 1/2 } ]' + [ x^{ -3 } ]' \cdot ( 1 - e^{ 2x } )^{ 1/2 }. #

# "Using the Chain Rule twice on the first differentiated quantity:" #

# g'(x) \ = #

# \ \ \ x^{ -3 } \cdot [ 1/2 ( 1 - e^{ 2x } )^{ -1/2 } ( 1 - e^{ 2x } )' ] + [ -3 x^{ -4 } ] \cdot ( 1 - e^{ 2x } )^{ 1/2 } #

# \qquad \qquad \ = #

# \ \ \ x^{ -3 } \cdot [ 1/2 ( 1 - e^{ 2x } )^{ -1/2 } ( 0 - 2 e^{ 2x } ) ] + ( -3 x^{ -4 } ) \cdot ( 1 - e^{ 2x } )^{ 1/2 }. #

# "Simplify:" #

# \qquad \qquad \ = #

# \ \ \ x^{ -3 } \cdot [ 1/2 ( 1 - e^{ 2x } )^{ -1/2 } ( - 2 e^{ 2x } ) ] + ( -3 x^{ -4 } ) \cdot ( 1 - e^{ 2x } )^{ 1/2 } #

# \qquad \qquad \ = #

# \ \ \ x^{ -3 } \cdot [ - ( 1 - e^{ 2x } )^{ -1/2 } ( e^{ 2x } ) ] + ( -3 x^{ -4 } ) \cdot ( 1 - e^{ 2x } )^{ 1/2 }. #

# "Continue simplification by pulling out lowest powers of same" #
# "quantities:" #

# \qquad \qquad \ = #

# \ \ \ - x^{ -4 } ( 1 - e^{ 2x } )^{ -1/2 } \cdot ( - x^1 ( e^{ 2x } ) + ( -3 ) \cdot ( 1 - e^{ 2x } )^{ 1 } ) #

# \qquad \qquad \ = \ \ \ x^{ -4 } ( 1 - e^{ 2x } )^{ -1/2 } \cdot [ ( x e^{ 2x } ) + 3\cdot ( 1 - e^{ 2x } ) ] #

# \qquad \qquad \ = \ \ \ x^{ -4 } ( 1 - e^{ 2x } )^{ -1/2 } \cdot [ x e^{ 2x } + 3 - 3 e^{ 2x } ] #

# \qquad \qquad \ = \ \ \ x^{ -4 } ( 1 - e^{ 2x } )^{ -1/2 } \cdot [ 3 + x e^{ 2x } - 3 e^{ 2x } ] #

# \qquad \qquad \ = \ \ \ x^{ -4 } ( 1 - e^{ 2x } )^{ -1/2 } \cdot [ 3 + ( x- 3 ) e^{ 2x } ]. #

# "Remove negative exponents, write one quantity as a square root:" #

# \qquad \qquad \ = \ \ \ [ 3 + ( x- 3 ) e^{ 2x } ] / { x^{ 4 } \sqrt{ 1 - e^{ 2x } } } \quad. #

#"This is our answer." #

# \ #

# "Summarizing:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad g(x) \ = \ x^{ -3 } \cdot ( 1 - e^{ 2x } )^{ 1/2 }. #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad g'(x) \ = \ [ 3 + ( x- 3 ) e^{ 2x } ] / { x^{ 4 } \sqrt{ 1 - e^{ 2x } } } \quad. #