Chapter 4 - Calculating the Derivative - 4.2 Derivatives of Products and Quotients - 4.2 Exercises - Page 216: 37

To sketch the method used to prove the product rule, let $$ f(x)=\frac{u(x)}{v(x)} $$ Then $$ f(x+h)=\frac{u(x+h)}{v(x+h)} $$ and, by definition, $f^{\prime}(x)$ is given by $$ \begin{aligned} f^{\prime}(x) &=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\frac{u(x+h)}{v(x+h)}-\frac{u(x)}{v(x)}}{h}\\ &=\lim _{h \rightarrow 0} \frac{u(x+h) v(x)-u(x) v(x+h)}{h v(x+h) v(x)} \\ &=\lim _{h \rightarrow 0} \frac{u(x+h) v(x)-u(x) v(x)+u(x) v(x)-u(x) v(x+h)}{h v(x+h) v(x)} \\ &=\lim _{h \rightarrow 0} \frac{v(x)[u(x+h)-u(x)]-u(x)[v(x+h)-v(x)]}{h v(x+h) v(x)} \\ &=\lim _{h \rightarrow 0} \frac{v(x) \frac{u(x+h)-u(x)}{h}-u(x) \frac{v(x+h)-v(x)}{h}}{v(x+h) v(x)} \\ &=\frac{v(x) \cdot u^{\prime}(x)-u(x) v^{\prime}(x)}{[v(x)]^{2}} \end{aligned} $$

To sketch the method used to prove the product rule, let $$ f(x)=\frac{u(x)}{v(x)} $$ Then $$ f(x+h)=\frac{u(x+h)}{v(x+h)} $$ and, by definition, $f^{\prime}(x)$ is given by $$ \begin{aligned} f^{\prime}(x) &=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\frac{u(x+h)}{v(x+h)}-\frac{u(x)}{v(x)}}{h}\\ &=\lim _{h \rightarrow 0} \frac{u(x+h) v(x)-u(x) v(x+h)}{h v(x+h) v(x)} \\ &=\lim _{h \rightarrow 0} \frac{u(x+h) v(x)-u(x) v(x)+u(x) v(x)-u(x) v(x+h)}{h v(x+h) v(x)} \\ &=\lim _{h \rightarrow 0} \frac{v(x)[u(x+h)-u(x)]-u(x)[v(x+h)-v(x)]}{h v(x+h) v(x)} \\ &=\lim _{h \rightarrow 0} \frac{v(x) \frac{u(x+h)-u(x)}{h}-u(x) \frac{v(x+h)-v(x)}{h}}{v(x+h) v(x)} \\ &=\frac{v(x) \cdot u^{\prime}(x)-u(x) v^{\prime}(x)}{[v(x)]^{2}} \end{aligned} $$