Question

Find dy/dx of log (1+√x)^2

Answer 1

y=2 [log1*(1/2)logx]
dy=log1*1/x *dx
dy/dx=0

Answer 2

log (1+√x)^2 = ln (1+√x)^2/ln 10
We'll differentiate with respect to x:
[ln (1+√x)^2/ln 10]' = (1/ln 10)*(1/(1+√x)^2)*[(1+√x)^2]'
[ln (1+√x)^2/ln 10]' = 2*(1+√x)*(1+√x)' /(ln 10)*(1+√x)^2
[ln (1+√x)^2/ln 10]' = 2*(1+√x)/2√x*(ln 10)*(1+√x)^2
dy/dx = [ln (1+√x)^2/ln 10]' =(1+√x) /√x*(ln 10)*(1+√x)^2