Exercise 5.2.3

Question

\begin{enumerate}[label=(\alph*)] 
  \item Use Definition 5.2.1 to produce the proper formula for the derivative of $h(x)=1 / x$.
  \item Combine the result in part (a) with the Chain Rule (Theorem 5.2.5) to supply a proof for part (iv) of Theorem 5.2.4.
  \item Supply a direct proof of Theorem 5.2.4 (iv) by algebraically manipulating the difference quotient for $(f / g)$ in a style similar to the proof of Theorem 5.2.4 (iii).
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item
    $$h'(x) = \lim_{y	o x} \frac{1/y - 1/x}{y-x} = \lim_{y 	o x} \frac{x - y}{xy(y-x)} = \lim_{y 	o x} -\frac{1}{xy} = -\frac{1}{x^2}$$
  \item By chain rule $(1/g)'(x) = -g'(x)/g(x)^2$ combined with product rule gives
    $$
    \begin{aligned}
      (f \cdot 1/g)'(x) = f'(x) \cdot \frac{1}{g(x)} + f(x)(1/g)'(x) &= \frac{f'(x)}{g(x)} - f(x)\frac{g'(x)}{g(x)^2} \
                                                                     &= \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}
    \end{aligned}
    $$
  \item
    $$
    (f/g)'(x) = \lim_{y 	o x} \frac{(f/g)(y) - (f/g)(x)}{y - x} = \lim_{y 	o x} \frac{f(y)g(x) - f(x)g(y)}{g(x)g(y)(y-x)}
    $$
    Now the $g(x)g(y)$ goes to $g(x)^2$, we just need to evaluate the derivatives in the numerator. We do this by adding and subtracting $f(x)g(x)$ similar to the proof of the product rule, then use the functional limit theorem to finish it off
    $$
    \lim_{y 	o x} \frac{g(x)(f(y)- f(x)) + f(x)(g(x) - g(y))}{g(x)g(y)(y-x)} = \frac{g(x)f'(x) - f(x)g'(x)}{g(x)^2}
    $$
   \end{enumerate}

Exercise 5.2.3

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Exercise 5.2.3

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