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