How do you differentiate #y = x ^ sqrt(x)#?

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Answer 1 · 2016-10-05T19:08:45

#dy/dx=1/2 x^(-1/2 + sqrt[x]) (2 + Log_e(x))#

#y = x ^ sqrt(x)#

Applying the #log# transformation ti both sides

#log_ey=sqrt(x) log_e x# so

#dy/y = (1/2log_e x/sqrt(x)+sqrt(x)/x)dx# so

#dy/dx = (1/2log_e x/sqrt(x)+sqrt(x)/x)y = (1/2log_e x/sqrt(x)+sqrt(x)/x)x^sqrt(x)#

Finally

#dy/dx=1/2 x^(-1/2 + sqrt[x]) (2 + Log_e(x))#