Find the derivative of 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\,=\,\ln \frac{(x+5)(x-1)}{x}.}
| Foundations:
|
| This problem is best approached through properties of logarithms. Remember that
|
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 \ln (xy) = \ln x + \ln y,}
|
| 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 \ln \left( \frac{x}{y}\right) = \ln x - \ln y,}
|
| 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
|
- 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 \left(\ln x\right)'\,=\,\frac{1}{x}.}
|
|
Solution:
| Step 1:
|
| We can use the log rules to rewrite our function as
|
- 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\,=\,\ln (x+5)+\ln(x-1)-\ln(x).}
|
| Step 2:
|
| We can differentiate term-by-term, applying the chain rule to the first two to find
|
- 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 \begin{array}{rcl} y' & = & \displaystyle{\frac{1}{x+5}\cdot(x+5)'+\frac{1}{x-1}\cdot(x-1)'+\frac{1}{x}}\\ \\ & = & \displaystyle{\frac{1}{x+5}+\frac{1}{x-1}+\frac{1}{x}}. \end{array}}
|
|
| Final Answer:
|
|
Return to Sample Exam