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anonymous
 one year ago
How do I differentiate (200000ln(t0.1))/(39.95t^2)
anonymous
 one year ago
How do I differentiate (200000ln(t0.1))/(39.95t^2)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I thought Quotient rule was needed but I couldn't work it out

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have to find the first derivative of this function: \[\Large f\left( t \right) = \frac{{{K_1}}}{{{K_2}}}\frac{{\ln \left( {t  a} \right)}}{{{t^2}}}\] where: \[\Large \begin{gathered} {K_1} = 200,000 \hfill \\ {K_2} = 39.95 \hfill \\ a = 0.1 \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1as remarked by @saseal we can apply the quotient rule to this function: \[\Large \frac{{\ln \left( {t  a} \right)}}{{{t^2}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1hint: if we have a quotient between these two functions: \[\Large \frac{{f\left( x \right)}}{{g\left( x \right)}}\] then the first derivative of such quotient, is: \[\Large \frac{{f'\left( x \right)g\left( x \right)  f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\] where f' (x) and g '(x) stand for first derivative of f(x) and g(x) respectively

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now you have to compute the first derivative of: ln(ta) and t^2 what functions do you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0u = ln(t  0.1) du/dt = 1/t0.1 v=t^2 dv/dt = 2t Is that right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now we can write this ratio: \[\Large \begin{gathered} \frac{{f'\left( x \right)g\left( x \right)  f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{\frac{1}{{t  a}}{t^2}  \ln \left( {t  a} \right)2t}}{{{t^4}}} = ...? \hfill \\ \end{gathered} \] please simplify

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1it is a simple algebraic computation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1hint: \[\Large \begin{gathered} \frac{{f'\left( x \right)g\left( x \right)  f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{\frac{1}{{t  a}}{t^2}  \ln \left( {t  a} \right)2t}}{{{t^4}}} = \hfill \\ \hfill \\ = \frac{{\frac{{{t^2}  \left( {t  a} \right)\ln \left( {t  a} \right)2t}}{{t  a}}}}{{{t^4}}} = ...? \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1which can be simplified to this expression: \[\Large \begin{gathered} \frac{{f'\left( x \right)g\left( x \right)  f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{\frac{1}{{t  a}}{t^2}  \ln \left( {t  a} \right)2t}}{{{t^4}}} = \hfill \\ \hfill \\ = \frac{{\frac{{{t^2}  \left( {t  a} \right)\left\{ {\ln \left( {t  a} \right)} \right\}2t}}{{t  a}}}}{{{t^4}}} = \hfill \\ \hfill \\ = \frac{{{t^2}  \left( {t  a} \right)\left\{ {\ln \left( {t  a} \right)} \right\}2t}}{{\left( {t  a} \right){t^4}}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now multiply that expression by \[\Large \frac{{{K_1}}}{{{K_2}}}\] and you will find the requested first derivative
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