1.

differentiate: log(logx)

Answer» Let `y="log"("log"x)`
`implies(dy)/(dx)=(d)/(dx)"log"("log"x)`
`=(1)/("log"x)(d)/(dx)("log"x)`
`=(1)/(x"log"x)=(x log x)^(-1)`
`implies(d^(2)y)/(dx^(2))=(d)/(dx)(x log x)^(-1)`
`= -1* (x logx)^(-2)(d)/(dx)(x logx)`
`=([x*(d)/(dx)logx+"log"x(d)/(dx)x])/((xlogx)^(2))`
`=([x*(1)/(x)+"log"x])/((xlogx)^(2))`
`=((1+"log"x))/((xlogx)^(2))`


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