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anonymous

  • one year ago

How do I differentiate (200000ln(t-0.1))/(39.95t^2)

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  1. anonymous
    • one year ago
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    quotient rule first

  2. anonymous
    • one year ago
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    I thought Quotient rule was needed but I couldn't work it out

  3. Michele_Laino
    • one year ago
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    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: \[\Large \begin{gathered} {K_1} = 200,000 \hfill \\ {K_2} = 39.95 \hfill \\ a = 0.1 \hfill \\ \end{gathered} \]

  4. Michele_Laino
    • one year ago
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    as remarked by @saseal we can apply the quotient rule to this function: \[\Large \frac{{\ln \left( {t - a} \right)}}{{{t^2}}}\]

  5. Michele_Laino
    • one year ago
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    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

  6. Michele_Laino
    • one year ago
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    now you have to compute the first derivative of: ln(t-a) and t^2 what functions do you get?

  7. anonymous
    • one year ago
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    u = ln(t - 0.1) du/dt = 1/t-0.1 v=t^2 dv/dt = 2t Is that right?

  8. Michele_Laino
    • one year ago
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    right!

  9. Michele_Laino
    • one year ago
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    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

  10. anonymous
    • one year ago
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    I'm not sure??

  11. Michele_Laino
    • one year ago
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    why?

  12. Michele_Laino
    • one year ago
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    it is a simple algebraic computation

  13. Michele_Laino
    • one year ago
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    hint: \[\Large \begin{gathered} \frac{{f'\left( x \right)g\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{\frac{1}{{t - a}}{t^2} - \ln \left( {t - a} \right)2t}}{{{t^4}}} = \hfill \\ \hfill \\ = \frac{{\frac{{{t^2} - \left( {t - a} \right)\ln \left( {t - a} \right)2t}}{{t - a}}}}{{{t^4}}} = ...? \hfill \\ \end{gathered} \]

  14. Michele_Laino
    • one year ago
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    which can be simplified to this expression: \[\Large \begin{gathered} \frac{{f'\left( x \right)g\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{\frac{1}{{t - a}}{t^2} - \ln \left( {t - a} \right)2t}}{{{t^4}}} = \hfill \\ \hfill \\ = \frac{{\frac{{{t^2} - \left( {t - a} \right)\left\{ {\ln \left( {t - a} \right)} \right\}2t}}{{t - a}}}}{{{t^4}}} = \hfill \\ \hfill \\ = \frac{{{t^2} - \left( {t - a} \right)\left\{ {\ln \left( {t - a} \right)} \right\}2t}}{{\left( {t - a} \right){t^4}}} \hfill \\ \end{gathered} \]

  15. Michele_Laino
    • one year ago
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    now multiply that expression by \[\Large \frac{{{K_1}}}{{{K_2}}}\] and you will find the requested first derivative

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