L02E03.tex

\documentclass[solutions.tex]{subfiles}

\xtitle

\begin{document}
\maketitle
\begin{exercise} Use the chain rule to find the derivatives
on each of the following functions:
\begin{equation*} \begin{aligned}
	g_0(t) &&=\quad& \sin(t^2) - \cos(t^2) \\
	\theta_0(\alpha) &&=\quad& e^{3\alpha} + 3\alpha\ln(3\alpha) \\
	x_0(t) &&=\quad& \sin^2(t^2) - \cos(t^2)
\end{aligned} \end{equation*}
\end{exercise}
\begin{remark} We've slightly altered the functions names compared
to what they have in the book. The reason will be apparent very soon.
\end{remark}
\begin{remark} As in previous exercises (e.g.
\href{https://github.com/mbivert/ttm/blob/master/cm/L02E01.pdf}{L02E01}),
we'll be using some additional mathematical notation to avoid us
the need to define functions. The meaning should be obvious from the
context. \\

We will also use the prime notation to denote differentiation, as we also did
in earlier exercises.
\end{remark}
There are two main ways of proceeding: either applying the \textit{chain
rule} to each individual term of each function, for instance to
compute $g'$, we would fist compute $(\sin(t^2))' = (\sin((u \mapsto u^2)(t)))'$,
using the chain rule, then $(\cos(t^2))'$ in a similar fashion, etc. \\

Or, and we're going to use this approach as it's likely to have been
expected one, we can consider each function "globally" as
a composition of two smaller functions. While doing so, we will find
that $g_0$, $\theta_0$ and $x_0$ are respectively created by composing
functions $f$, $\theta$ and $x$ of an earlier exercise, with two simple
functions $(v \mapsto v^2)$ and $(v \mapsto 3v)$:
\begin{equation*} \begin{aligned}
	g_0(t) &&=\quad&
		(\underbrace{(u \mapsto \sin(u) - \cos(u))}_g\circ(v \mapsto v^2))(t) \\
	\theta_0(\alpha) &&=\quad&
		(\underbrace{(u \mapsto e^{u} + u\ln(u))}_\theta\circ(v \mapsto 3v))(t) \\
	x_0(t) &&=\quad&
		(\underbrace{(u \mapsto \sin^2(u) - \cos(u))}_x\circ(v \mapsto v^2))(t)
\end{aligned} \end{equation*}

Let's remember the derivative of $g$, $\theta$ and $x$, that we've
computed in
\href{https://github.com/mbivert/ttm/blob/master/cm/L02E01.pdf}{L02E01}:
\begin{equation*} \begin{aligned}
	g'(x) &&=\quad& \cos(x)+\sin(x) \\
	\theta'(\alpha) &&=\quad& e^\alpha+\ln(\alpha)+1 \\
	x'(t) &&=\quad& (1+2\cos t)\sin t
\end{aligned} \end{equation*}

Finally, let's recall the \textit{chain rule}:
\[
	\frac{d}{dt}(\psi\circ\varphi) = \frac{d}{dt}(\psi(\varphi(t)) = \varphi'(t)\psi'(\varphi(t))
\]

Then, the derivative of $(v \mapsto v^2)$ and $(v \mapsto 3v)$ being respectively
$(v \mapsto 2v)$ and $(v \mapsto 3)$ (constant function), we have the following
derivatives for our functions:
\begin{equation*} \begin{aligned}
	g_0'(t) &&=\quad& \boxed{2t(\cos t^2 + \sin t^2)} \\
	\theta_0'(\alpha) &&=\quad& \boxed{3(e^{3\alpha}+\ln(3\alpha)+1)} \\
	x_0'(t) &&=\quad& \boxed{2t((1+2\cos t^2)\sin t^2)}
\end{aligned} \end{equation*}

\end{document}