anonymous
  • anonymous
How do I differentiate (200000ln(t-0.1))/(39.95t^2)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
quotient rule first
anonymous
  • anonymous
I thought Quotient rule was needed but I couldn't work it out
Michele_Laino
  • Michele_Laino
we 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: \[\Large \begin{gathered} {K_1} = 200,000 \hfill \\ {K_2} = 39.95 \hfill \\ a = 0.1 \hfill \\ \end{gathered} \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Michele_Laino
  • Michele_Laino
as remarked by @saseal we can apply the quotient rule to this function: \[\Large \frac{{\ln \left( {t - a} \right)}}{{{t^2}}}\]
Michele_Laino
  • Michele_Laino
hint: 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
  • Michele_Laino
now you have to compute the first derivative of: ln(t-a) and t^2 what functions do you get?
anonymous
  • anonymous
u = ln(t - 0.1) du/dt = 1/t-0.1 v=t^2 dv/dt = 2t Is that right?
Michele_Laino
  • Michele_Laino
right!
Michele_Laino
  • Michele_Laino
now 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
anonymous
  • anonymous
I'm not sure??
Michele_Laino
  • Michele_Laino
why?
Michele_Laino
  • Michele_Laino
it is a simple algebraic computation
Michele_Laino
  • Michele_Laino
hint: \[\Large \begin{gathered} \frac{{f'\left( x \right)g\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{\frac{1}{{t - a}}{t^2} - \ln \left( {t - a} \right)2t}}{{{t^4}}} = \hfill \\ \hfill \\ = \frac{{\frac{{{t^2} - \left( {t - a} \right)\ln \left( {t - a} \right)2t}}{{t - a}}}}{{{t^4}}} = ...? \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
which 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
  • Michele_Laino
now multiply that expression by \[\Large \frac{{{K_1}}}{{{K_2}}}\] and you will find the requested first derivative

Looking for something else?

Not the answer you are looking for? Search for more explanations.