How do you differentiate # f(x)=e^sqrt(3lnx+x^2)# using the chain rule.?

1 Answer
Jan 16, 2016

Answer:

#dy/dx=(e^sqrt(3lnx+x^2) * (3/x+2x))/((2sqrt(3lnx+x^2)) #

Explanation:

The chain rule:

#dy/dx=dy/(du) * (du)/(dv)*(dv)/(dx)#

#y = e^u, dy/(du)=e^u#

#u=v^(1/2), (du)/(dv)=1/2v^-(1/2)#

#v=3lnx+x^2, (dv)/dx=3/x+2x#

#dy/dx=e^u*1/(2sqrtv)*(3/x+2x)#

#dy/dx=(e^sqrt(3lnx+x^2) * (3/x+2x))/((2sqrt(3lnx+x^2)) #