Chain Rule for One-Dimensional Derivatives

Let $\mathbb{R}$ be the real numbers, and let $|\cdot|$ denote the absolute value on $\mathbb{R}$.

Let $I$ and $J$ be intervals, let $\gamma:I\to\mathbb{R}$ satisfy $\gamma(t)\in J$ for every $t\in I$, and let $g:J\to\mathbb{R}$. Let $g\circ\gamma:I\to\mathbb{R}$ be the function whose value at $t\in I$ is $g(\gamma(t))$.

Let $t_0\in I$ be an interior point of $I$ at which $\gamma$ is differentiable, and assume that $\gamma(t_0)$ is an interior point of $J$ at which $g$ is differentiable.

Then $g\circ\gamma$ is differentiable at $t_0$, and

$$(g\circ\gamma)'(t_0)=g'\bigl(\gamma(t_0)\bigr)\,\gamma'(t_0).$$