Question

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ y=\left(\frac{6-5 x}{x^{2}-1}\right)^{2} $$

Short answer

The derivative of the function is \(y'=2(\frac{6-5x}{x^2-1}) \cdot (\frac{-5(x^2-1)-(6-5x)(2x)}{(x^2-1)^2})\)

Step-by-step solution

Apply Chain Rule

To handle the square of the whole fraction, use the chain rule. The chain rule states that the derivative of \((f(g(x)))') = f'(g(x)) \cdot g'(x)\). Here, consider \(f(u) = u^2\) and \(u= g(x) = \frac{6-5x}{x^2-1}\). So, the derivative of \(y\) at this step becomes: \(y' = 2u \cdot u'\), where \(u'\) is the derivative of \(u\) with respect to \(x\), which needs to be computed using the quotient rule.

Apply Quotient Rule

To find \(u'\), apply the quotient rule to \(\frac{6-5x}{x^2-1}\). The quotient rule is defined as: \((\frac{f}{g})' = \frac{f'g - fg'}{g^2}\). Here, \(f = 6-5x\) and \(g = x^2 - 1\). From this, we can find that \(f' = -5\) and \(g' = 2x\). Thus, we can compute \(u'\) as : \(u' = \frac{f'g - fg'}{g^2} = \frac{-5(x^2 - 1) - (6-5x)(2x)}{(x^2 - 1)^2}\)

Substitute \(u'\) Back

Now we need to substitute \(u'\) back into the derivative found in step 1, to get the final derivative of our original function \(y\). So the final derivative \(y'\) becomes: \(y'=2(\frac{6-5x}{x^2-1}) \cdot (\frac{-5(x^2-1)-(6-5x)(2x)}{(x^2-1)^2})\)