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

1 Answer
Feb 13, 2018

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

Explanation:

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"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."

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"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.