How do you find the derivatives of #y=x^(2x)# by logarithmic differentiation?

Question

38172 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-11T15:34:19

The answer is #=2(1+lnx)x^(2x)#

We need
#(uv)'=u'v+uv'#

#y=x^(2x)#

#lny=ln(x^(2x))#

#lny=2xlnx#

Differentiating wrt #x#

#1/ydy/dx=2(x*1/x+1*lnx)#

#dy/dx=2(1+lnx)y#

#dy/dx=2(1+lnx)x^(2x)#