How would one calculate absolute error under root operations? I know absolutely nothing about error analysis and I was wondering if someone could help me out. I want to approximate the value of $t$ that satisfies this equation: $$\sqrt{(\frac{1}{200}e^{2t}-\frac{1}{200}e^{-2t})^{2}+(\frac{1}{50}e^{2t})^{2}}=1.$$ I want to drop the term with the negative power, to get the following:
\begin{align}
\sqrt{(\frac{1}{200}e^{2t})^{2}+(\frac{1}{50}e^{2t})^{2}}&=1 \\
e^{2t}\sqrt{(\frac{1}{200})^{2}+(\frac{1}{50})^{2}}&=1 \\
2t+\frac{1}{2}\ln{\left((\frac{1}{200})^{2}+(\frac{1}{50})^{2}\right)
}&=0 \\
t&=-\frac{1}{4}\ln\left((\frac{1}{200})^{2}+(\frac{1}{50})^{2}\right)
\end{align}
However, I don't know by how much this approximates my actual value of t. How would one go about calculating the absolute error? Thanks in advance.

I'll not answer the error propagation part, but show you how to calculate the $t$ value explicitly: By squaring the inital equation and setting $X = e^{2t}$, you obtain
$$\frac{1}{200^2}(X-\frac{1}{X})^2 + \frac{1}{50^2}X^2 = 1$$
$$\frac{1}{200}(X^2 - 2 + \frac{1}{X^2}) + \frac{4}{50}X^2 = 200$$
$$\frac{1}{200}(X^4 - 2X^2 + 1) + \frac{4}{50}X^4 = 200 X^2$$
$$(X^4 - 2X^2 +1) + 16 X^4 = 40000 X^2$$
$$17X^4 - 40002X^2 + 1 = 0$$
Now substitute $Y = X^2$, solve the quadratic equation, take the square root to obtain $X$ and then the logarithm to obtain $t$.
