Question

# If $$f\left( x \right)$$ and $$g\left( x \right)$$ are two functions with $$g\left( x \right) =x-\dfrac { 1 }{ x }$$ and $$f\circ g\left( x \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } }$$, then $$f^{ ' }\left( x \right)$$ is equal to

A
3x2+3
B
x21x2
C
1+1x2
D
3x2+3x4

Solution

## The correct option is A $$3{ x }^{ 2 }+3$$$$f\circ g\left( x \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } }$$Writing $${ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } }$$ using $${ \left( a-b \right) }^{ 3 }={ a }^{ 3 }-{ b }^{ 3 }-3ab\left( a-b \right)$$, we have$${ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } -3x\cdot \dfrac { 1 }{ x } \left( x-\dfrac { 1 }{ x } \right)$$$$\Rightarrow { x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right)$$We have,$$f\left( g\left( x \right) \right) ={ x }^{ 3 }-\dfrac { 1 }{ { x }^{ 3 } } ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right)$$As $$g\left( x \right) =x-\dfrac { 1 }{ x }$$, this yields$$f\left( x-\dfrac { 1 }{ x } \right) ={ \left( x-\dfrac { 1 }{ x } \right) }^{ 3 }+3\left( x-\dfrac { 1 }{ x } \right)$$On putting $$x-\dfrac { 1 }{ x } =t$$, we get$$f\left( t \right) ={ t }^{ 3 }+3t$$Thus, $$f\left( x \right) ={ x }^{ 3 }+3x$$and $$f^{ ' }\left( x \right) =3{ x }^{ 2 }+3$$Mathematics

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