What is the second derivative of #f(x)= xe^(x^3)#?

1 Answer
Jan 16, 2016

#f''(x)=3(3x^3+4)x^2e^(x^3)#

Explanation:

#f''(x)=d/(dx)f'(x)#

#f(x)=h(x)*e^(g(x))#

#h(x)=x;#

#g(x)=x^3#

We have to apply Product Rule and Chain Rule
Let's start founding #f'(x)#

#f'(x)=h'(x)e^(g(x))+h(x)*[d/(dx)e^g(x)*d/dxg(x)]=#

#=1*e^(x^3)+x*[e^(x^3)*3x^2]=(1+3x^3)e^(x^3)#

Now we can computate #f''(x)# applying again Product Rule and Chain Rule

#f''(x)=[d/dx(1+3x^3)]e^(x^3)+(1+3x^3)*[d/dxe^(x^3)d/dxx^3]=#

#=[0+3*3x^2]e^x+(1+3x^3)[e^(x^3)*3x^2]=#

#=3*3x^2e^(x^3)+(1+3x^3)*3x^2e^(x^3)=#

#=3x^2e^(x^3)(3+1+3x^3)=#

#=3(3x^3+4)x^2e^(x^3)#