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

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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