find the derivative of the function. y = ((x^2 + 1)/(x^2 - 1))^7 y =

Explanation:

Step1: Apply chain - rule

Let $u=\frac{x^{2}+1}{x^{2}-1}$, then $y = u^{7}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$. Since $y = u^{7}$, $\frac{dy}{du}=7u^{6}=7(\frac{x^{2}+1}{x^{2}-1})^{6}$.

Step2: Apply quotient - rule to find $\frac{du}{dx}$

The quotient - rule states that if $u=\frac{f(x)}{g(x)}$ where $f(x)=x^{2}+1$ and $g(x)=x^{2}-1$, then $\frac{du}{dx}=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g^{2}(x)}$. Here, $f^{\prime}(x) = 2x$ and $g^{\prime}(x)=2x$. So $\frac{du}{dx}=\frac{2x(x^{2}-1)-2x(x^{2}+1)}{(x^{2}-1)^{2}}=\frac{2x^{3}-2x - 2x^{3}-2x}{(x^{2}-1)^{2}}=\frac{-4x}{(x^{2}-1)^{2}}$.

Step3: Calculate $\frac{dy}{dx}$

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}$ and $\frac{du}{dx}$: $\frac{dy}{dx}=7(\frac{x^{2}+1}{x^{2}-1})^{6}\cdot\frac{-4x}{(x^{2}-1)^{2}}=\frac{-28x(x^{2}+1)^{6}}{(x^{2}-1)^{8}}$.

Answer:

$\frac{-28x(x^{2}+1)^{6}}{(x^{2}-1)^{8}}$