Find the derivative of the function. $$ y=\cos \left(\frac{1-e^{3 x}}{1+e^{3 x}}\right) $$ $$ y^{\prime}=\square $$

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Let 
$$
u(x)=\frac{1-e^{3x}}{1+e^{3x}}.
$$
Then
$$
y=\cos\bigl(u(x)\bigr)
$$
and by the chain rule,
$$
y'=-\sin\bigl(u(x)\bigr)\cdot u'(x).
$$

Now, we differentiate $$ u(x) $$. Write
$$
u(x)=\frac{1-e^{3x}}{1+e^{3x}}.
$$
Using the quotient rule, where for $$ \frac{f(x)}{g(x)} $$ we have 
$$
\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2},
$$
here $$ f(x)=1-e^{3x} $$ and $$ g(x)=1+e^{3x} $$.

Differentiate $$ f(x) $$ and $$ g(x) $$:
$$
f'(x)=-3e^{3x},\quad g'(x)=3e^{3x}.
$$

Thus,
$$
u'(x)=\frac{-3e^{3x}(1+e^{3x}) - (1-e^{3x})(3e^{3x})}{(1+e^{3x})^2}.
$$
Simplify the numerator:
$$
-3e^{3x}(1+e^{3x}) - 3e^{3x}(1-e^{3x}) = -3e^{3x}\Bigl[(1+e^{3x})+(1-e^{3x})\Bigr] = -3e^{3x}\cdot 2 = -6e^{3x}.
$$
So,
$$
u'(x) = \frac{-6e^{3x}}{(1+e^{3x})^2}.
$$

Substitute $$u'(x)$$ back into the derivative of $$y$$:
$$
y'=-\sin \left(\frac{1-e^{3x}}{1+e^{3x}}\right) \cdot \left(\frac{-6e^{3x}}{(1+e^{3x})^2}\right).
$$
The two negatives cancel, giving:
$$
y'=\frac{6e^{3x}}{(1+e^{3x})^2}\sin\left(\frac{1-e^{3x}}{1+e^{3x}}\right).
$$

Thus,
$$
y'=\frac{6e^{3x}}{(1+e^{3x})^2}\sin\left(\frac{1-e^{3x}}{1+e^{3x}}\right).
$$
The derivative of the function $$ y=\cos \left(\frac{1-e^{3x}}{1+e^{3x}}\right) $$ is: $$ y'=\frac{6e^{3x}}{(1+e^{3x})^2}\sin\left(\frac{1-e^{3x}}{1+e^{3x}}\right). $$

Find the derivative of the function.

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