Question

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

Answer 1

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 }`

Answer 2

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