How do you differentiate #y=e^(2x^3)#?

1 Answer
Oct 5, 2016

#(dy)/(dx)=6x^2e^(2x^3)#

Explanation:

We use Chain Rule here.

In order to differentiate a function of a function, say #y, =f(g(x))#, where we have to find #(dy)/(dx)#, we need to do substitute #u=g(x)#, which gives us #y=f(u)#.

The Chain Rule states that #(dy)/(dx)=(dy)/(du)xx(du)/(dx)#.

In fact if we have something like #y=f(g(h(x)))#, we can have #(dy)/(dx)=(dy)/(df)xx(df)/(dg)xx(dg)/(dh)#

Here we have #y=e^(2x^3)#

Hence #(dy)/(dx)=(d(e^(2x^3)))/(d(2x^3))xx(d(2x^3))/(dx)#

= #e^(2x^3)xx2xx3x^2#

= #6x^2e^(2x^3)#