 $\vec g$ be functions so that the composition $\vec f(\vec
    g(\vec x))$ makes sense (we can use the output of $g$ as an
    input to $f$). Suppose $\vec f$ is differentiable at $\vec
    g(\vec x)$ and that $\vec g$ is differentiable at $\vec x$.
    Then the derivative of $\vec f\circ \vec g$ at $\vec x$ is
    $$D(\vec f\circ \vec g)(\vec x) = D\vec f(\vec g(\vec x))\cdot
    D\vec g(\vec x).$$ The derivative of a composition is equal to
    the derivative of the outside (evaluated at the inside),
   