Difference between revisions of "022 Exam 2 Sample B, Problem 1"
Jump to navigation
Jump to search
| Line 40: | Line 40: | ||
|<br> | |<br> | ||
::<math>\begin{array}{rcl} | ::<math>\begin{array}{rcl} | ||
| − | y | + | y & = & \displaystyle{\ln \frac{(x+1)^4}{(2x - 5)(x + 4)}}\\ |
\\ | \\ | ||
& = & 4\ln (x+1)-\ln(2x-5)-\ln (x+4). | & = & 4\ln (x+1)-\ln(2x-5)-\ln (x+4). | ||
| Line 47: | Line 47: | ||
| | | | ||
|- | |- | ||
| + | |<br> | ||
|- | |- | ||
|} | |} | ||
Latest revision as of 21:22, 20 January 2017
Find the derivative of
| Foundations: | |
|---|---|
| This problem is best approached through properties of logarithms. Remember that | |
| while | |
| and | |
| You will also need to apply | |
| The Chain Rule: If 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} 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 g} are differentiable functions, then | |
| 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\circ g)'(x) = f'(g(x))\cdot g'(x).} | |
| Finally, recall that the derivative of natural log is | |
|
Solution:
| Step 1: | |
|---|---|
| We can use the log rules to rewrite our function as | |
|
|
| Step 2: | |
|---|---|
| We can differentiate term-by-term, applying the chain rule to each term to find | |
|
| Final Answer: |
|---|
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 y'\,=\,\displaystyle{\frac{4}{x+1}-\frac{2}{2x-5}-\frac{1}{x+4}}.} |