Earlier we saw that $y(x) = e^{3x^2-1}$ is a composite function that can be written as $y(x) = u(v(x))$, with $v(x) = 3x^2 -1$ and $u(v) = e^v$. By the use of the chain rule we can determine the derivative of this function:
$$
\begin{align}
v'(x) &= 3\cdot 2x = 6x\\
u'(v) &= e^v\\
y'(x) &= u'\big(v(x)\big)\cdot v'(x) = e^{v(x)} \cdot 6x = 6xe^{3x^2-1}.
\end{align}
$$
$$
\begin{align}
v'(x) &= 3\cdot 2x = 6x\\
u'(v) &= e^v\\
y'(x) &= u'\big(v(x)\big)\cdot v'(x) = e^{v(x)} \cdot 6x = 6xe^{3x^2-1}.
\end{align}
$$