Answer
$$t '(x)= \frac{t \cos (x t)}{1-x \cos (x t)} $$
Work Step by Step
Given $$\sin (x t)=t$$
Differentiate with respect to $x$
\begin{align*}
\frac{d}{dx} \sin (x t)&=\frac{d}{dx} t \\
\left( xt'(x)+t \right ) \cos (x t)&= t'(x) \\
\left (1-x\cos (x t) \right) t'(x)& =-t \cos (x t)
\end{align*}
Then
$$t '(x)= \frac{t \cos (x t)}{1-x \cos (x t)} $$
