Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 90

Find a formula for $\frac{d^3}{dx^3}$ having $y = f(u)$ and $u = g(x)$ where $f$ and $g$ possess third derivatives. Recall that $\displaystyle \frac{d^2y}{dx^2} = \frac{d^2y}{du^2} \left( \frac{du}{dx} \right)^2 + \frac{dy}{du} \left( \frac{d^2u}{dx^2} \right) $

Using Chain Rule,

$\displaystyle \frac{d^3y}{dx^2} = \left[ \frac{d}{dx} \left( \frac{d^2y}{du^2} \right) \cdot \left( \frac{du}{dx} \right)^2 \right] + \left[ \frac{d^2y}{du^2} \cdot \frac{d}{dx} \left( \frac{du}{dx} \right)^2 \right] + \left[ \frac{d}{dx} \left( \frac{d}{du} \right) \cdot \frac{d^2u}{dx^2} \right] + \left[ \frac{dy}{du} \cdot \frac{d}{dx} \left( \frac{d^2u}{dx^2} \right) \right]$

For the first bracket,

$\displaystyle \frac{d}{dx} \left( \frac{d^2y}{du^2} \right) = \frac{d}{du} \left( \frac{d^2y}{du^2} \right) \left( \frac{du}{dx} \right) = \frac{d^3y}{du^3} \left( \frac{du}{dx} \right)$

For the second bracket,

$\displaystyle \frac{d}{dx} \left( \frac{du}{dx} \right) ^2 = 2 \left( \frac{du}{dx} \right)' \cdot \frac{d}{dx} \left( \frac{du}{dx} \right) = 2 \frac{du}{dx} \left( \frac{d^2 u}{dx^2} \right)$

For the third bracket,

$\displaystyle \frac{d}{dx} \left( \frac{dy}{du} \right) = \frac{d}{du} \left( \frac{dy}{du} \right) \left( \frac{du}{dx} \right) = \frac{d^2y}{du^2 } \left( \frac{du}{dx} \right)$

For the fourth bracket,

$\displaystyle \frac{d}{dx} \left( \frac{d^2 u}{dx^2} \right) = \frac{d^3u}{dx^3}$

Therefore,


$
\begin{equation}
\begin{aligned}

\frac{d^3y}{dx^3} =& \left[ \frac{d^3y}{du^3} \left( \frac{du}{dx} \right) \cdot \left( \frac{du}{dx}\right)^2 \right] +
\left[ \frac{d^2y}{du^2} \cdot 2 \frac{du}{dx} \left( \frac{d^2 u}{dx^2} \right) \right] +
\left[ \frac{d^2y}{du^2} \left( \frac{du}{dx} \right) \cdot \frac{d^2u}{dx^2} \right] +
\left[ \frac{dy}{du} \cdot \frac{d^3 u}{dx^3} \right]
\\
\\
\\
\\
\frac{d^3y}{dx^3} =& \frac{d^3y}{du^3} \left( \frac{du}{dx} \right) ^3 + 2 \frac{d^2y}{du^2} \frac{du}{dx} \left( \frac{d^2 u}{dx^2} \right) + \frac{d^2 y}{du^2} \frac{du}{dx} \left( \frac{d^2 u}{dx^2} \right) + \frac{dy}{du} \frac{d^3u}{dx^3}
\\
\\
\\
\\
\frac{d^3y}{dx^3} =& \frac{d^3 y}{du^3} \left( \frac{du}{dx} \right)^3 + 3 \frac{d^2y}{du^2} \frac{du}{dx} \frac{d^2 u}{dx^2} + \frac{dy}{du} \frac{d^3u}{dx^3}

\end{aligned}
\end{equation}
$