Question

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

Answer 1

quotient rule first

Answer 2

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

Answer 3

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:
\[\Large \begin{gathered}
{K_1} = 200,000 \hfill \\
{K_2} = 39.95 \hfill \\
a = 0.1 \hfill \\
\end{gathered} \]

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Is that right?

Answer 8

right!

Answer 9

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

Answer 10

I'm not sure??

Answer 11

why?

Answer 12

it is a simple algebraic computation

Answer 13

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

Answer 14

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

Answer 15

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