Difference between revisions of "022 Exam 1 Sample A, Problem 1"

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|We can't plug in <math style="vertical-align: 0%">h=0</math>, as this would lead to the unallowed "division by zero". Instead, we multiply by the conjugate of the numerator and clean up:
 
|We can't plug in <math style="vertical-align: 0%">h=0</math>, as this would lead to the unallowed "division by zero". Instead, we multiply by the conjugate of the numerator and clean up:
 
<table>
 
<table>
<tr>
+
  <tr>
<td>&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{h\rightarrow0}\frac{\sqrt{x+h-5}-\sqrt{x-5}}{h}\cdot\frac{\sqrt{x+h-5}+\sqrt{x-5}}{\sqrt{x+h-5}+\sqrt{x-5}}</math></td>
+
    <td>&nbsp;&nbsp;&nbsp;&nbsp; <math>\lim_{h\rightarrow0}\frac{\sqrt{x+h-5}-\sqrt{x-5}}{h}\cdot\frac{\sqrt{x+h-5}+\sqrt{x-5}}{\sqrt{x+h-5}+\sqrt{x-5}}</math></td>
<td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{\left(\sqrt{x+h-5}\right)^{2}-\left(\sqrt{x-5}\right)^{2}}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
+
    <td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{\left(\sqrt{x+h-5}\right)^{2}-\left(\sqrt{x-5}\right)^{2}}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
</tr>
+
  </tr>
<tr>
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  <tr>
<td></td>
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    <td></td>
<td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{x+h-5-\left(x-5\right)}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math> </td>
+
    <td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{x+h-5-\left(x-5\right)}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math> </td>
</tr>
+
  </tr>
<tr>
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  <tr>
<td></td>
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    <td></td>
<td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{h}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
+
    <td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{h}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
</tr>
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  </tr>
<tr>
+
  <tr>
<td></td>
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    <td></td>
<td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{1}{\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
+
    <td>&nbsp; <math>=\,\,\lim_{h\rightarrow0}\frac{1}{\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}</math></td>
<tr>
+
  <tr>
<td></td>
+
    <td></td>
<td>&nbsp; <math>=\,\,\frac{1}{2\sqrt{x-5}}.</math></td>
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    <td>&nbsp; <math>=\,\,\frac{1}{2\sqrt{x-5}}.</math></td>
 
</table>
 
</table>
 
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Revision as of 11:43, 2 April 2015

1. Use the definition of derivative to find the derivative of .

Foundations:  
Recall that the derivative is actually defined through the limit
     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}.}
The goal in solving this is to plug the values Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x+h} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} into the appropriate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} in the numerator, and then find a way to cancel the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h} in the denominator. Unlike simplifying a rational expression containing radicals, here it's appropriate to have a radical in the denominator.
Again, the goal is to cancel the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h} .

 Solution:

Step 1:  
Following the hints above, we initially have
     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)\,=\,\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\,=\,\lim_{h\rightarrow0}\frac{\sqrt{x+h-5}-\sqrt{x-5}}{h}.}
Step 2:  
We can't plug in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h=0} , as this would lead to the unallowed "division by zero". Instead, we multiply by the conjugate of the numerator and clean up:
     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{h\rightarrow0}\frac{\sqrt{x+h-5}-\sqrt{x-5}}{h}\cdot\frac{\sqrt{x+h-5}+\sqrt{x-5}}{\sqrt{x+h-5}+\sqrt{x-5}}}   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\,\,\lim_{h\rightarrow0}\frac{\left(\sqrt{x+h-5}\right)^{2}-\left(\sqrt{x-5}\right)^{2}}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}}
  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\,\,\lim_{h\rightarrow0}\frac{x+h-5-\left(x-5\right)}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}}
  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\,\,\lim_{h\rightarrow0}\frac{h}{h\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}}
  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\,\,\lim_{h\rightarrow0}\frac{1}{\left(\sqrt{x+h-5}+\sqrt{x-5}\right)}}
  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\,\,\frac{1}{2\sqrt{x-5}}.}
Notice that this is the same result you would get using our more convenient "rules of integration", including the chain rule, but that's not the point of this problem. You specifically need to treat the derivative as a limit.
Final Answer:  
Note that no points would be given for only a correct answer. This exam question is about correctly using the definition of a derivative. However, the correct answer is
     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)\,\,=\,\,\frac{1}{2\sqrt{x-5}}.}

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