Answer:
![(d)/(dx) (ln(e^(x^2)) ) = (1)/(e^(x^2)) 2x e^(x^2) = 2x](https://img.qammunity.org/2021/formulas/mathematics/college/le560da5oom8nuc1oe8t0z5vlyryxolz62.png)
Explanation:
For this case we want to find the derivate of this function:
![y = ln(e^(x^2)) )](https://img.qammunity.org/2021/formulas/mathematics/college/5tkxfp4zt7thpyxf3kt27iabr90g18t4vy.png)
And in order to find the derivate we need to apply the chain rule given by:
![(df(u))/(dx) =(df)/(du) (du)/(dx)](https://img.qammunity.org/2021/formulas/mathematics/college/5lctob7ls54tglvl2zkzzpn38wfs807yee.png)
And on this case
![f = ln(u), u = e^(x^2)](https://img.qammunity.org/2021/formulas/mathematics/college/fh7dctoxm05v2jfi2zfvbrnl3rvce7i7bj.png)
And we can find the partial derivates like this:
![(d)/(du) (ln(u)) =(1)/(u)](https://img.qammunity.org/2021/formulas/mathematics/college/vcymsd59qd02b5iuw3i5uesdg87yhloel4.png)
![(d)/(dx)(e^(x^2))= e^(x^2) (2x)](https://img.qammunity.org/2021/formulas/mathematics/college/g0o01r79olt62p76yd8l63irnk27fbl19e.png)
And if we replace we got:
![(d)/(dx) (ln(e^(x^2)) ) = (1)/(u) 2x e^(x^2)](https://img.qammunity.org/2021/formulas/mathematics/college/9re8aotpsrrbmj65joaio398m8i7p4i9lc.png)
And if we replace
we got:
![(d)/(dx) (ln(e^(x^2)) ) = (1)/(e^(x^2)) 2x e^(x^2) = 2x](https://img.qammunity.org/2021/formulas/mathematics/college/le560da5oom8nuc1oe8t0z5vlyryxolz62.png)
And that would be our final answer on this case