Question

dy/dx of -1/a (ln((a+bx)/x) ?

Answer 1

find dy/dx of
\[y = -\frac{ 1 }{ a }\ln (\frac{ a+bx }{ x })\]

Answer 2

a and b are constants

Answer 3

\[\frac{ dy }{ dx } = -\frac{ 1 }{ a }(\frac{ x }{ a+bx })(\frac{ bx-a-bx }{ x^2 })\]
is that right?

Answer 4

Nope, maybe if you can simplify more then the answer is \[\frac{1}{a x+b x^2}\]

Answer 5

i have the answer, i just need the work...
so my work was incorrect?

Answer 6

From product rule you get
\[\frac{dy}{dx}=-\frac{1}{a}(-\frac{a}{ax+bx^2})+\ln(\frac{a+bx}{x})\frac{1}{a^2}\]

Answer 7

ohhhhh i just figured it out.
nvm i think i got it.

Answer 8

There should be 2 terms

Answer 9

yeah i kinda figured after you mentioned it.