1. $$ \left(e^{x^{2}}\right)^{\prime \prime} $$

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We start with the function $$ e^{x^{2}} $$ and need to compute its second derivative, i.e., $$ \left(e^{x^{2}}\right)'' $$.
To compute the first derivative, we use the chain rule. Letting $$ u = x^{2} $$, then $$ e^{x^{2}} = e^{u} $$ and $$ \frac{du}{dx} = 2x $$. The derivative is:
$$
\frac{d}{dx} e^{x^{2}} = e^{x^{2}} \cdot 2x = 2x e^{x^{2}}.
$$
Now, differentiate the first derivative $$ 2x e^{x^{2}} $$ using the product rule. The product rule states that if $$ f(x) = u(x)v(x) $$, then $$ f'(x) = u'(x)v(x) + u(x)v'(x) $$.

Letting $$ u(x) = 2x $$ and $$ v(x) = e^{x^{2}} $$:

- Derivative of $$ u(x) $$ is $$ u'(x) = 2 $$.
- Derivative of $$ v(x) $$ is computed using the chain rule: $$ v'(x) = e^{x^{2}} \cdot 2x $$.

Thus, the second derivative becomes:
$$
\frac{d}{dx}(2x e^{x^{2}}) = 2 e^{x^{2}} + 2x \cdot (2x e^{x^{2}}) = 2 e^{x^{2}} + 4x^{2} e^{x^{2}}.
$$

We can factor out $$ e^{x^{2}} $$:
$$
(e^{x^{2}})'' = e^{x^{2}} (2 + 4x^{2}).
$$
The second derivative of $$ e^{x^{2}} $$ is $$ (e^{x^{2}})'' = e^{x^{2}} (2+4x^{2}) $$.

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