Quotient vs Product Rule
You are asked to differentiate $$ y = \frac{x - 1}{x + 1} $$ Looking at the question, I'm thinking I could solve this question using the product rule by making $\tfrac{1}{x + 1}$ into $(x + 1)^{-1}$. Is there something I am not conceptually understanding about the two rules?
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$\begingroup$The quotient rule and the product rule are the same thing. In particular, the quotient rule follows from the product rule and the chain rule. Recall that the product rule states that if $h(x) = f(x) g(x)$, then $$h'(x) = f'(x) g(x) + f(x) g'(x).$$ Also recall that the chain rule states that if $h(x) = f(g(x))$, then $$h'(x) = f'(g(x))g'(x).$$ Therefore, if $$h(x) = f(x)/g(x) = f(x)\cdot \frac{1}{g(x)},$$ then the product rule gives $$h'(x) = f'(x) \cdot \frac{1}{g(x)} + f(x) \frac{d}{dx}\left[\frac{1}{g(x)}\right].$$ Then the chain rule applied to the second term gives $$\frac{d}{dx}\left[\frac{1}{g(x)}\right] = -(g(x))^{-2} g'(x).$$ Therefore, $$h'(x) = \frac{f'(x)}{g(x)} - \frac{f(x)g'(x)}{g(x)^2} = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}.$$
$\endgroup$ $\begingroup$Sure, you can use either of the two rules: the quotient rule or the product rule. If you try each approach, you'll see that they each lead to the same derivative, $y'$.
To simplify the function even further, You could also use the fact that $$y = \frac{x-1}{x+1} = y = \frac{x +1 - 1 - 1}{x+1} = 1 - \frac{2}{x+1} = 1 -2(x+1)^{-1}$$
Then $$y' = (-1)(-2)(x+ 1)^{-2} = \dfrac 2{(x+1)^2}$$
$\endgroup$ $\begingroup$You can find the derivative using the product rule, provided that you also know the chain rule. Since $$ \begin{align} \frac{d}{dx} \left[ \frac{x - 1}{x + 1} \right] &= \frac{d}{dx} \left[ (x - 1) \cdot (x + 1)^{-1} \right] \\ &= \frac{d}{dx} (x - 1) \cdot (x + 1)^{-1} + (x - 1) \cdot \frac{d}{dx} \left[ (x + 1)^{-1} \right]. \end{align} $$ The last derivative is of the composite function $$ x \mapsto (x + 1) \mapsto (x + 1)^{-1}. $$
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