find the derivative of the function.\n$y = \\cos(\\frac{1 - e^{2x}}{1 + e^{2x}})$\n$y=$\n14. - / 1…

Explanation:

Step1: Apply chain - rule

Let $u=\frac{1 - e^{2x}}{1 + e^{2x}}$, then $y = \cos(u)$. The chain - rule states that $y'=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$. Since $y=\cos(u)$, $\frac{dy}{du}=-\sin(u)=-\sin(\frac{1 - e^{2x}}{1 + e^{2x}})$.

Step2: Find $\frac{du}{dx}$ using quotient - rule

The quotient - rule for $u=\frac{f(x)}{g(x)}$ where $f(x)=1 - e^{2x}$ and $g(x)=1 + e^{2x}$ is $\frac{du}{dx}=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$. We know that $f'(x)=-2e^{2x}$ and $g'(x)=2e^{2x}$. Then $\frac{du}{dx}=\frac{-2e^{2x}(1 + e^{2x})-(1 - e^{2x})\cdot2e^{2x}}{(1 + e^{2x})^2}=\frac{-2e^{2x}-2e^{4x}-2e^{2x}+2e^{4x}}{(1 + e^{2x})^2}=\frac{-4e^{2x}}{(1 + e^{2x})^2}$.

Step3: Calculate $y'$

By the chain - rule $y'=\frac{dy}{du}\cdot\frac{du}{dx}=-\sin(\frac{1 - e^{2x}}{1 + e^{2x}})\cdot\frac{-4e^{2x}}{(1 + e^{2x})^2}=\frac{4e^{2x}\sin(\frac{1 - e^{2x}}{1 + e^{2x}})}{(1 + e^{2x})^2}$.

Answer:

$\frac{4e^{2x}\sin(\frac{1 - e^{2x}}{1 + e^{2x}})}{(1 + e^{2x})^2}$