1. ## derivative of 1/ln

Hey guys, I can't seem to figure this problem out, can anyone help??
its:

y = 1 / ln (cx)

find derivate of y... and the value of c is found by rearranging the equation above and isolating c

I keep getting that c = 1 / xe^y
and then i get that the derivative is y' = e^y / (ln (1/e^y)^2 but i know its wrong because the answer should be: y' = -y^2 / x

2. Originally Posted by collegestudent321
Hey guys, I can't seem to figure this problem out, can anyone help??
its:

y = 1 / ln (cx)

find derivate of y... and the value of c is found by rearranging the equation above and isolating c

I keep getting that c = 1 / xe^y
and then i get that the derivative is y' = e^y / (ln (1/e^y)^2 but i know its wrong because the answer should be: y' = -y^2 / x
$y = [\ln(cx)]^{-1}$

$y' = -[\ln(cx)]^{-2} \cdot \frac{1}{x}$

$y' = -\left([\ln(cx)]^{-1}\right)^2 \cdot \frac{1}{x}$

$y' = -(y)^2 \cdot \frac{1}{x} = -\frac{y^2}{x}$

3. Originally Posted by collegestudent321
Hey guys, I can't seem to figure this problem out, can anyone help??
its:

y = 1 / ln (cx)

find derivate of y... and the value of c is found by rearranging the equation above and isolating c

I keep getting that c = 1 / xe^y
and then i get that the derivative is y' = e^y / (ln (1/e^y)^2 but i know its wrong because the answer should be: y' = -y^2 / x
do you know implicit differentiation ? if you do write as

yln(cx)=1

if not by quotient rule with

u=1 du=0
v=ln(cx) dv=1/x

$
y'= \frac {(\frac{-1}{x})}{(ln(cx))^2}
$

tidy this up and use

$
y=\frac{1}{ ln(cx)}
$

4. Originally Posted by collegestudent321
Hey guys, I can't seem to figure this problem out, can anyone help??
its:

y = 1 / ln (cx)

find derivate of y... and the value of c is found by rearranging the equation above and isolating c

I keep getting that c = 1 / xe^y
and then i get that the derivative is y' = e^y / (ln (1/e^y)^2 but i know its wrong because the answer should be: y' = -y^2 / x
$y=(\ln{(cx)})^{-1}$

$y'=-(\ln{(cx)})^{-2}(\frac{1}{cx})(c)$

$y'=-(y)^2\frac{1}{x}$

5. got it! makes perfect sense! thanx guys!