Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 22

Suppose that $g(x) + x \sin g(x) = x^2$, find $g'(0)$
Solving for $g(0)$

$\displaystyle\frac{d}{dx} [ g(x) ] + \frac{d}{dx} [x \sin g(x)] = \frac{d}{dx} (x^2)$

$
\begin{equation}
\begin{aligned}
\frac{d}{dx} [g(x)] + \left[ (x) \frac{d}{dx} [ \sin g(x)] + [ \sin g(x) ] \frac{d}{dx} (x) \right] &= \frac{d}{dx} (x^2)\\
\\
g'(x) + (x) [ \cos g(x) ] \frac{d}{dx} [ g(x) ] + [ \sin g(x)] (1) &= 2x\\
\\
g'(x) + (x) [g'(x)] [ \cos g(x)] + \sin g(x) &= 2x
\end{aligned}
\end{equation}
$


If $x = 0 $


$
\begin{equation}
\begin{aligned}
g(0) + (0) \sin g(0) &= (0)^2\\
\\
g(0) &= 0\\
\\
g'(0) + (0) [g'(0)] [\cos g(0)] + \sin g(0) &= 2(0)\\
\\
g'(0) + 0 + 0 &= 0\\
\\
g'(0) &= 0
\end{aligned}
\end{equation}
$