If `y=log_(10)x+log_(x)10+log_(x)x+log_(10)10` then what is `((dy)/(dx))_(x=10)` equal to?

Options

A. `(1)/(xlog_(e)10)-(log_(e)10)/(x(log_(e)x)^(2))`
B. `(1)/(xlog_(e)10)-(1)/(xlog_(10)e)`
C. `(1)/(xlog_(e)10)+(1)/(x(log_(e)x)^(2))`
D. None of the above

Answer

A

Explanation

Given
`y= log_(10)x+ log_(10)x 10+ log_(x)x+ log_(10)10`
`rArr y= log_(10) e.log_(e) x+(log_(e)10)/(log_(e)x)+1+1`
On differentiating w.r.t. x we get
`(dy)/(dx)=(1)/(x) log_(10)e- (log_(e)10)/(x (log_(e)x)^(2))`
`x=(1)/(x log_(e)10)-(log_(e)10)/(x (log_(e)x)^(2))`