question 10 (3 points)\nuse the first principles definition to determine the derivative of the general…

Explanation:

Step1: Recall the first - principles formula

The derivative of a function (y = f(x)) using the first - principles definition is (f^{\prime}(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}). Given (f(x)=\frac{1}{ax + b}), then (f(x + h)=\frac{1}{a(x + h)+b}=\frac{1}{ax+ah + b}).

Step2: Substitute into the formula

[ \begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{\frac{1}{ax+ah + b}-\frac{1}{ax + b}}{h}\ &=\frac{\frac{(ax + b)-(ax+ah + b)}{(ax + b)(ax+ah + b)}}{h}\ &=\frac{\frac{ax + b-ax - ah - b}{(ax + b)(ax+ah + b)}}{h}\ &=\frac{\frac{-ah}{(ax + b)(ax+ah + b)}}{h} \end{align*} ]

Step3: Simplify the expression

Cancel out the (h) terms: (\frac{-ah}{h(ax + b)(ax+ah + b)}=\frac{-a}{(ax + b)(ax+ah + b)})

Step4: Take the limit as (h\rightarrow0)

(\lim_{h\rightarrow0}\frac{-a}{(ax + b)(ax+ah + b)}). As (h\rightarrow0), (ax+ah + b\rightarrow ax + b). So (f^{\prime}(x)=\frac{-a}{(ax + b)^2})

Answer:

(f^{\prime}(x)=\frac{-a}{(ax + b)^2})