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

Question

Ezra Adams · Accepted Answer

# (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.!

Isaiah Allen · Answer

undefined To find the first and second derivatives of $ y = x^{e^{cx}} $:

First derivative:
$$ \frac{dy}{dx} = e^{cx} \cdot x^{e^{cx} - 1} \cdot (1 + c \cdot \ln(x)) $$

Second derivative:
$$ \frac{d^2y}{dx^2} = e^{cx} \cdot x^{e^{cx} - 1} \cdot (1 + c \cdot \ln(x))^2 + e^{2cx} \cdot x^{e^{cx} - 2} \cdot c $$

These derivatives can be obtained using the chain rule and product rule of differentiation.