Difference between revisions of "022 Exam 1 Sample A, Problem 8"

Latest revision as of 15:34, 2 April 2015

8. Find the derivative of the function $f(x)={\frac {(3x-1)^{2}}{x^{3}-7}}$ . You do not need to simplify your answer.

Foundations:
This problem involves some more advanced rules of differentiation. In particular, it requires
The Chain Rule: If $f$ and $g$ are differentiable functions, then
$(f\circ g)'(x)=f'(g(x))\cdot g'(x).$
The Quotient Rule: If $f$ and $g$ are differentiable functions and $g(x)\neq 0$ , then
$\left({\frac {f}{g}}\right)'(x)={\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^{2}}}.$

Solution:

Note that we need to use chain rule to find the derivative of

\left(3x-1\right)^{2}

. Then we find

$f'(x)$	$=\,\,{\frac {\left[\left(3x-1\right)^{2}\right]'\cdot (x^{3}-7)\,\,-\,\,\left(3x-1\right)^{2}\cdot (x^{3}-7)'}{(x^{3}-7)^{2}}}$
	$=\,\,{\frac {\left[2\left(3x-1\right)\cdot 3\right]\cdot (x^{3}-7)\,\,-\,\,\left(3x-1\right)^{2}\cdot 3x^{2}}{(x^{3}-7)^{2}}}.$

@@ Line 26: / Line 26: @@
 |<table>
    <tr style="vertical-align: middle">
-     <td> &nbsp;&nbsp;&nbsp;&nbsp; <math style="vertical-align: -70%">f'(x)</math>&nbsp; </td>
+     <td> &nbsp;&nbsp;&nbsp;&nbsp; <math style="vertical-align: -70%">f'(x)</math>&nbsp;&nbsp; </td>
      <td><math>=\,\,\frac{\left[\left(3x-1\right)^{2}\right]'\cdot(x^{3}-7) \,\,-\,\, \left(3x-1\right)^{2}\cdot(x^{3}-7)'}{(x^{3}-7)^{2}}</math></td>
    </tr>
    <tr>
      <td></td>
-     <td><math>=\frac{\left[2\left(3x-1\right)\cdot3\right]\cdot(x^{3}-7) \,\,-\,\, \left(3x-1\right)^{2}\cdot3x^{2}}{(x^{3}-7)^{2}}.</math></td>
+     <td><math>=\,\,\frac{\left[2\left(3x-1\right)\cdot3\right]\cdot(x^{3}-7) \,\,-\,\, \left(3x-1\right)^{2}\cdot3x^{2}}{(x^{3}-7)^{2}}.</math></td>
 </table>
 |}