## anonymous one year ago How do I differentiate (200000ln(t-0.1))/(39.95t^2)

• This Question is Open
1. anonymous

quotient rule first

2. anonymous

I thought Quotient rule was needed but I couldn't work it out

3. 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: $\Large \begin{gathered} {K_1} = 200,000 \hfill \\ {K_2} = 39.95 \hfill \\ a = 0.1 \hfill \\ \end{gathered}$

4. Michele_Laino

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

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

now you have to compute the first derivative of: ln(t-a) and t^2 what functions do you get?

7. anonymous

u = ln(t - 0.1) du/dt = 1/t-0.1 v=t^2 dv/dt = 2t Is that right?

8. Michele_Laino

right!

9. 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

10. anonymous

I'm not sure??

11. Michele_Laino

why?

12. Michele_Laino

it is a simple algebraic computation

13. Michele_Laino

hint: $\Large \begin{gathered} \frac{{f'\left( x \right)g\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{\frac{1}{{t - a}}{t^2} - \ln \left( {t - a} \right)2t}}{{{t^4}}} = \hfill \\ \hfill \\ = \frac{{\frac{{{t^2} - \left( {t - a} \right)\ln \left( {t - a} \right)2t}}{{t - a}}}}{{{t^4}}} = ...? \hfill \\ \end{gathered}$

14. Michele_Laino

which can be simplified to this expression: $\Large \begin{gathered} \frac{{f'\left( x \right)g\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{\frac{1}{{t - a}}{t^2} - \ln \left( {t - a} \right)2t}}{{{t^4}}} = \hfill \\ \hfill \\ = \frac{{\frac{{{t^2} - \left( {t - a} \right)\left\{ {\ln \left( {t - a} \right)} \right\}2t}}{{t - a}}}}{{{t^4}}} = \hfill \\ \hfill \\ = \frac{{{t^2} - \left( {t - a} \right)\left\{ {\ln \left( {t - a} \right)} \right\}2t}}{{\left( {t - a} \right){t^4}}} \hfill \\ \end{gathered}$

15. Michele_Laino

now multiply that expression by $\Large \frac{{{K_1}}}{{{K_2}}}$ and you will find the requested first derivative