find the derivative of the function. y = (e^4u - e^-4u)/(e^4u + e^-4u) y = need help? read it submit answer…

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{f(u)}{g(u)}$, then $y'=\frac{f'(u)g(u)-f(u)g'(u)}{g(u)^2}$. Here, $f(u)=e^{4u}-e^{-4u}$ and $g(u)=e^{4u}+e^{-4u}$.

Step2: Find $f'(u)$ and $g'(u)$

Using the chain - rule, if $y = e^{au}$, then $y'=ae^{au}$. So, $f'(u)=4e^{4u}+4e^{-4u}$ and $g'(u)=4e^{4u}-4e^{-4u}$.

Step3: Apply the quotient - rule

[ \begin{align*} y'&=\frac{(4e^{4u}+4e^{-4u})(e^{4u}+e^{-4u})-(e^{4u}-e^{-4u})(4e^{4u}-4e^{-4u})}{(e^{4u}+e^{-4u})^2}\ &=\frac{4(e^{4u}+e^{-4u})^2-4(e^{4u}-e^{-4u})^2}{(e^{4u}+e^{-4u})^2}\ &=\frac{4\left[(e^{4u}+e^{-4u})^2-(e^{4u}-e^{-4u})^2\right]}{(e^{4u}+e^{-4u})^2}\ \end{align*} ] Using the difference of squares formula $(a + b)^2-(a - b)^2=(a^2 + 2ab + b^2)-(a^2-2ab + b^2)=4ab$, with $a = e^{4u}$ and $b = e^{-4u}$, we have: [ \begin{align*} y'&=\frac{4\times4e^{4u}e^{-4u}}{(e^{4u}+e^{-4u})^2}\ &=\frac{16}{(e^{4u}+e^{-4u})^2} \end{align*} ]

Answer:

$\frac{16}{(e^{4u}+e^{-4u})^2}$