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Calculus
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Use the limit definition of the derivative to find \(\displaystyle f'(x)\) if \(\displaystyle f(x)= 4 x^{2} + x - 5\).

Using the difference quotient \(\displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\) gives \(\displaystyle \lim_{h\to 0}\frac{(h + x + 4 \left(h + x\right)^{2} - 5)-(4 x^{2} + x - 5)}{h}\) Expanding and collecting like terms gives \(\displaystyle \lim_{h \to 0}\frac{4 h^{2} + 8 h x + h}{h}\) Factoring out \(\displaystyle h\) and reducing gives \(\displaystyle \lim_{h\to 0}4 h + 8 x + 1\). Evaluating the limit gives \(\displaystyle f'(x)=8 x + 1\)

Download \(\LaTeX\) 

\begin{question}Use the limit definition of the derivative to find $f'(x)$ if $f(x)= 4 x^{2} + x - 5$. 
    \soln{9cm}{Using the difference quotient $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$ gives $\lim_{h\to 0}\frac{(h + x + 4 \left(h + x\right)^{2} - 5)-(4 x^{2} + x - 5)}{h}$ Expanding and collecting like terms gives $\lim_{h \to 0}\frac{4 h^{2} + 8 h x + h}{h}$ Factoring out $h$ and reducing gives $\lim_{h\to 0}4 h + 8 x + 1$. Evaluating the limit gives $f'(x)=8 x + 1$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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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)= 4 x^{2} + x - 5 " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%3D%204%20x%5E%7B2%7D%20%2B%20x%20-%205%20" alt="LaTeX:  \displaystyle f(x)= 4 x^{2} + x - 5 " data-equation-content=" \displaystyle f(x)= 4 x^{2} + x - 5 " /> . </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{(h + x + 4 \left(h + x\right)^{2} - 5)-(4 x^{2} + x - 5)}{h} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bh%5Cto%200%7D%5Cfrac%7B%28h%20%2B%20x%20%2B%204%20%5Cleft%28h%20%2B%20x%5Cright%29%5E%7B2%7D%20-%205%29-%284%20x%5E%7B2%7D%20%2B%20x%20-%205%29%7D%7Bh%7D%20" alt="LaTeX:  \displaystyle \lim_{h\to 0}\frac{(h + x + 4 \left(h + x\right)^{2} - 5)-(4 x^{2} + x - 5)}{h} " data-equation-content=" \displaystyle \lim_{h\to 0}\frac{(h + x + 4 \left(h + x\right)^{2} - 5)-(4 x^{2} + x - 5)}{h} " />  Expanding and collecting like terms gives  <img class="equation_image" title=" \displaystyle \lim_{h \to 0}\frac{4 h^{2} + 8 h x + h}{h} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bh%20%5Cto%200%7D%5Cfrac%7B4%20h%5E%7B2%7D%20%2B%208%20h%20x%20%2B%20h%7D%7Bh%7D%20" alt="LaTeX:  \displaystyle \lim_{h \to 0}\frac{4 h^{2} + 8 h x + h}{h} " data-equation-content=" \displaystyle \lim_{h \to 0}\frac{4 h^{2} + 8 h x + h}{h} " />  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}4 h + 8 x + 1 " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bh%5Cto%200%7D4%20h%20%2B%208%20x%20%2B%201%20" alt="LaTeX:  \displaystyle \lim_{h\to 0}4 h + 8 x + 1 " data-equation-content=" \displaystyle \lim_{h\to 0}4 h + 8 x + 1 " /> . Evaluating the limit gives  <img class="equation_image" title=" \displaystyle f'(x)=8 x + 1 " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%3D8%20x%20%2B%201%20" alt="LaTeX:  \displaystyle f'(x)=8 x + 1 " data-equation-content=" \displaystyle f'(x)=8 x + 1 " /> </p> </p>