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

1 Answer
Aug 10, 2017

(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}.

Explanation:

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.!