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Question

If $$y=\log _{ 10 }{ x } +\log _{ x }{ 10 } +\log _{ x }{ x } +\log _{ 10 }{ 10 } $$, then $$\dfrac { dy }{ dx } $$ is equal to


A
1xloge10loge10x(logex)2
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B
1xloge101xlog10e
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C
1xloge10+loge10x(logex)2
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D
None of the above
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Solution

The correct option is B $$\dfrac { 1 }{ x\log _{ e }{ 10 } } -\dfrac { \log _{ e }{ 10 } }{ x{ \left( \log _{ e }{ x } \right) }^{ 2 } } $$
Given,
$$y=\log _{ 10 }{ x } +\log _{ x }{ 10 } +\log _{ x }{ x } +\log _{ 10 }{ 10 } $$
$$\Rightarrow y=\log _{ 10 }{ e } \cdot \log _{ e }{ x } +\dfrac { \log _{ e }{ 10 }  }{ \log _{ e }{ x }  } +1+1$$
On differentiating with respect to $$x$$, we get
$$\dfrac { dy }{ dx } =\dfrac { 1 }{ x } \log _{ 10 }{ e } -\dfrac { \log _{ e }{ 10 }  }{ x{ \left( \log _{ e }{ x }  \right)  }^{ 2 } } $$
    $$=\dfrac { 1 }{ x\log _{ e }{ 10 }  } -\dfrac { \log _{ e }{ 10 }  }{ x{ \left( \log _{ e }{ x }  \right)  }^{ 2 } } $$

Mathematics

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