If #y = log_10(x) + log_x(10) + log_x(x) + log_10(10)#, then #dy/dx =#?

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Answer 1 · 2018-05-02T05:37:08

#dy/dx = 1/(xln(10))-ln(10)/(((ln(x))^2)x)#

Given: #y = log_10(x) + log_x(10) + log_x(x) + log_10(10)#

Convert to base e:

#y = ln(x)/ln(10) + ln(10)/ln(x) + ln(x)/ln(x) + ln(10)/ln(10)#

Simplify:

#y = ln(x)/ln(10) + ln(10)/ln(x) + 1 + 1#

#y = (1/ln(10))ln(x)+ ln(10) (ln(x))^-1+2#

#dy/dx = 1/(xln(10))-ln(10)((ln(x))^-2)/x#

#dy/dx = 1/(xln(10))-ln(10)/(((ln(x))^2)x)#