« Product and Quotient Rule for Derivatives 

If $y = f(u)$ is a differentiable function of $u$, and $u = g(x)$ is a differentiable function of $x$, then $y = f(g(x))$. This a differentiable function of $x$, and
$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$or, equivalently,
$$\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)$$Example 1:
$$ \begin{aligned} y &= (x^2 + 1)^3, \ \frac{dy}{dx} = ? \\ g(x) &= u = x^2 + 1, \\ y &= f(u) = u^3, \\ \frac{df}{dx} &= \frac{df}{du} \cdot \frac{du}{dx} = (3u^2) (2x) = 3 (x^2 + 1)^2 (2x) = 6x (x^2 + 1)^2 \end{aligned} $$