Question

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

Answer 1

$f(x)=\dfrac{200000\ln(t-0.1)}{39.95t^2}=\dfrac{4000000}{799}\cdot\dfrac{\ln(t-0.1)}{t^2}\\ f'(x)=\dfrac{4000000}{799}\cdot\dfrac{\dfrac{1}{t-0.1}\cdot t^2-\ln(t-0.1)\cdot2t}{t^4}\\ f'(x)=\dfrac{4000000}{799}\cdot\dfrac{\dfrac{t}{t-0.1}-2\ln(t-0.1)}{t^3}\\f'(x)=\dfrac{4000000\left(\dfrac{t}{t-0.1}-2\ln(t-0.1)\right)}{799t^3}$

Answer 2

Answer:

$\frac { dy }{ dt } =\frac { 8\cdot { 10 }^{ 6 } }{ 799{ t }^{ 2 } } \left\{ \frac { 5 }{ 10t-1 } -\frac { \ln { \left( t-\frac { 1 }{ 10 } \right) } }{ t } \right\}$

Workings below. Don't know if it could've have been simplified further.

I've made a "smooth criminal" version, just in case you like things compressed.

$\frac { dy }{ dt } =\frac { n }{ { t }^{ 2 } } \left\{ \frac { 1 }{ t-k } -\frac { \ln { \left( { \left( t-k \right) }^{ 2 } \right) } }{ t } \right\} \\ \\ n=\frac { 4\cdot { 10 }^{ 6 } }{ 799 } ,\quad k=\frac { 1 }{ 10 }$