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Using the difference quotient \(\displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\) gives \(\displaystyle \lim_{h\to 0}\frac{\frac{9}{h + x + 1}-\frac{9}{x + 1}}{h}\) Using the least common denominator, \(\displaystyle (h + x + 1)(x + 1) \), to clear the fractions gives \(\displaystyle \frac{9(x + 1)-9(h + x + 1)}{h(x + 1)(h + x + 1)}\). Expanding gives \(\displaystyle \frac{- 9 h}{h \left(x + 1\right) \left(h + x + 1\right)}\). Factoring out \(\displaystyle h\) and reducing gives \(\displaystyle \lim_{h\to 0}- \frac{9}{\left(x + 1\right) \left(h + x + 1\right)}\). Evaluating the limit gives \(\displaystyle f'(x)=- \frac{9}{\left(x + 1\right)^{2}}\)

Download \(\LaTeX\)

\begin{question}Use the limit definition of the derivative to find $f'(x)$ if $f(x)= \frac{9}{x + 1}$. 
    \soln{9cm}{Using the difference quotient $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$ gives $\lim_{h\to 0}\frac{\frac{9}{h + x + 1}-\frac{9}{x + 1}}{h}$ Using the least common denominator, $(h + x + 1)(x + 1) $, to clear the fractions gives $\frac{9(x + 1)-9(h + x + 1)}{h(x + 1)(h + x + 1)}$.  Expanding gives $\frac{- 9 h}{h \left(x + 1\right) \left(h + x + 1\right)}$. Factoring out $h$ and reducing gives $\lim_{h\to 0}- \frac{9}{\left(x + 1\right) \left(h + x + 1\right)}$. Evaluating the limit gives $f'(x)=- \frac{9}{\left(x + 1\right)^{2}}$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

\end{question}\end{document}

HTML for Canvas

<p> <p>Use the limit definition of the derivative to find  <img class="equation_image" title=" \displaystyle f'(x) " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20" alt="LaTeX:  \displaystyle f'(x) " data-equation-content=" \displaystyle f'(x) " />  if  <img class="equation_image" title=" \displaystyle f(x)= \frac{9}{x + 1} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%3D%20%5Cfrac%7B9%7D%7Bx%20%2B%201%7D%20" alt="LaTeX:  \displaystyle f(x)= \frac{9}{x + 1} " data-equation-content=" \displaystyle f(x)= \frac{9}{x + 1} " /> . </p> </p>

HTML for Canvas

<p> <p>Using the difference quotient  <img class="equation_image" title=" \displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bh%20%5Cto%200%7D%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D%20" alt="LaTeX:  \displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} " data-equation-content=" \displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} " />  gives  <img class="equation_image" title=" \displaystyle \lim_{h\to 0}\frac{\frac{9}{h + x + 1}-\frac{9}{x + 1}}{h} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bh%5Cto%200%7D%5Cfrac%7B%5Cfrac%7B9%7D%7Bh%20%2B%20x%20%2B%201%7D-%5Cfrac%7B9%7D%7Bx%20%2B%201%7D%7D%7Bh%7D%20" alt="LaTeX:  \displaystyle \lim_{h\to 0}\frac{\frac{9}{h + x + 1}-\frac{9}{x + 1}}{h} " data-equation-content=" \displaystyle \lim_{h\to 0}\frac{\frac{9}{h + x + 1}-\frac{9}{x + 1}}{h} " />  Using the least common denominator,  <img class="equation_image" title=" \displaystyle (h + x + 1)(x + 1)  " src="/equation_images/%20%5Cdisplaystyle%20%28h%20%2B%20x%20%2B%201%29%28x%20%2B%201%29%20%20" alt="LaTeX:  \displaystyle (h + x + 1)(x + 1)  " data-equation-content=" \displaystyle (h + x + 1)(x + 1)  " /> , to clear the fractions gives  <img class="equation_image" title=" \displaystyle \frac{9(x + 1)-9(h + x + 1)}{h(x + 1)(h + x + 1)} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B9%28x%20%2B%201%29-9%28h%20%2B%20x%20%2B%201%29%7D%7Bh%28x%20%2B%201%29%28h%20%2B%20x%20%2B%201%29%7D%20" alt="LaTeX:  \displaystyle \frac{9(x + 1)-9(h + x + 1)}{h(x + 1)(h + x + 1)} " data-equation-content=" \displaystyle \frac{9(x + 1)-9(h + x + 1)}{h(x + 1)(h + x + 1)} " /> .  Expanding gives  <img class="equation_image" title=" \displaystyle \frac{- 9 h}{h \left(x + 1\right) \left(h + x + 1\right)} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B-%209%20h%7D%7Bh%20%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%28h%20%2B%20x%20%2B%201%5Cright%29%7D%20" alt="LaTeX:  \displaystyle \frac{- 9 h}{h \left(x + 1\right) \left(h + x + 1\right)} " data-equation-content=" \displaystyle \frac{- 9 h}{h \left(x + 1\right) \left(h + x + 1\right)} " /> . Factoring out  <img class="equation_image" title=" \displaystyle h " src="/equation_images/%20%5Cdisplaystyle%20h%20" alt="LaTeX:  \displaystyle h " data-equation-content=" \displaystyle h " />  and reducing gives  <img class="equation_image" title=" \displaystyle \lim_{h\to 0}- \frac{9}{\left(x + 1\right) \left(h + x + 1\right)} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bh%5Cto%200%7D-%20%5Cfrac%7B9%7D%7B%5Cleft%28x%20%2B%201%5Cright%29%20%5Cleft%28h%20%2B%20x%20%2B%201%5Cright%29%7D%20" alt="LaTeX:  \displaystyle \lim_{h\to 0}- \frac{9}{\left(x + 1\right) \left(h + x + 1\right)} " data-equation-content=" \displaystyle \lim_{h\to 0}- \frac{9}{\left(x + 1\right) \left(h + x + 1\right)} " /> . Evaluating the limit gives  <img class="equation_image" title=" \displaystyle f'(x)=- \frac{9}{\left(x + 1\right)^{2}} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%3D-%20%5Cfrac%7B9%7D%7B%5Cleft%28x%20%2B%201%5Cright%29%5E%7B2%7D%7D%20" alt="LaTeX:  \displaystyle f'(x)=- \frac{9}{\left(x + 1\right)^{2}} " data-equation-content=" \displaystyle f'(x)=- \frac{9}{\left(x + 1\right)^{2}} " /> </p> </p>