How do you find the first and second derivative of #y=x^(e^cx)#?

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Answer 1 · 2017-08-10T06:59:33

# (1) : dy/dx=e^cy(1+lnx)=e^cx^(e^c*x)(1+lnx).#

# (2) : (d^2y)/dx^2=e^cx^(e^c*x-1){1+xe^c(1+lnx)^2}.#

#y=x^(e^c*x)#

#:. ln y=ln{x^(e^c*x)}=(e^c*x)(lnx).#

#:. d/dx{lny}=d/dx{(e^c*x)(lnx)}.#

Applying the Chain Rule for the L.H.S., and, the

Product Rule for the R.H.S., we get,

#{d/dy(lny)}{dy/dx}=(xe^c)d/dx(lnx)+(lnx)d/dx((xe^c)).#

#:. 1/ydy/dx=(xe^c)(1/x)+(lnx)(e^cd/dx(x)).#

#=e^c+(e^c)(lnx)(1)=e^c(1+lnx)#

# rArr dy/dx=e^cy(1+lnx)..............(d_1).#

Next, #(d^2y)/dx^2=d/dx{dy/dx},#

#=d/dx{e^cy(1+lnx)},#

#=e^cd/dx{y(1+lnx)},#

#=e^c[yd/dx(1+lnx)+(1+lnx)d/dx(y)],#

#=e^c{y*1/x+(1+lnx)e^cy(1+lnx)}....[because, (d_1)],#

#=y/x*e^c{1+xe^c(1+lnx)^2}.#

Since, #y=x^(e^c*x),# we have,

#(d^2y)/dx^2=e^cx^(e^c*x-1){1+xe^c(1+lnx)^2}.#

Enjoy Maths.!