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)