1. ## logarithmic differentiation

use logarithmic differential to find dy/dx, where
$\displaystyle y = \frac{(x-4)^{-3}(\sin(x))^{4}}{(x^{2} - 2x)^{2} e^{x^{3}}}$

$\displaystyle y' = y*$________

sofar:
$\displaystyle ln y = ln\frac{(x-4)^{-3}(\sin(x))^{4}}{(x^{2} - 2x)^{2} e^{x^{3}}}$

= $\displaystyle -3ln(x-4) + 4ln(sin(x)) - 2ln(x^2+2x) + x^3$

= $\displaystyle \frac{1}{y}y' = \frac{-3}{x-4} + \frac{4}{cos(x)} -2\frac{2x+2}{x^2+2x} + 3x^2$

not sure what to do next

2. Originally Posted by viet
use logarithmic differential to find dy/dx, where
$\displaystyle y = \frac{(x-4)^{-3}(\sin(x))^{4}}{(x^{2} - 2x)^{2} e^{x^{3}}}$

$\displaystyle y' = y*$________

sofar:
$\displaystyle ln y = ln\frac{(x-4)^{-3}(\sin(x))^{4}}{(x^{2} - 2x)^{2} e^{x^{3}}}$

= $\displaystyle -3ln(x-4) + 4ln(sin(x)) - 2ln(x^2+2x) + x^3$

= $\displaystyle \frac{1}{y}y' = \frac{-3}{x-4} + \frac{4}{cos(x)} -2\frac{2x+2}{x^2+2x} + 3x^2$

not sure what to do next
first off, that is incorrect. (your differentiated $\displaystyle \ln (\sin x )$ incorrectly)

when you get the correct solution, just multiply both sides by y. you can replace y with its function of x if you wish

EDIT: oh, you dont have to write y as a function of x, look at the way they expect the answer, y*_______. so it would be y times whatever you got the derivative to be

3. i think this is correct so far:

$\displaystyle ln y = ln\frac{(x-4)^{-3}(\sin(x))^{4}}{(x^{2} - 2x)^{2} e^{x^{3}}}$

= $\displaystyle -3ln(x-4) + 4ln(sin(x)) - 2ln(x^2+2x) + x^3$

= $\displaystyle \frac{1}{y}y' = \frac{-3}{x-4} + \frac{4cos(x)}{sin(x)} -2\frac{2x+2}{x^2+2x} + 3x^2$

= $\displaystyle y\frac{-3}{x-4} + \frac{4cos(x)}{sin(x)} -\frac{4x+4}{x^2+2x} + 3x^2$

4. Originally Posted by viet
i think this is correct so far:

$\displaystyle ln y = ln\frac{(x-4)^{-3}(\sin(x))^{4}}{(x^{2} - 2x)^{2} e^{x^{3}}}$

= $\displaystyle -3ln(x-4) + 4ln(sin(x)) - 2ln(x^2+2x) + x^3$

= $\displaystyle \frac{1}{y}y' = \frac{-3}{x-4} + \frac{4cos(x)}{sin(x)} -2\frac{2x+2}{x^2+2x} + 3x^2$

= $\displaystyle y\frac{-3}{x-4} + \frac{4cos(x)}{sin(x)} -\frac{4x+4}{x^2+2x} + 3x^2$
no it isn't. it's y times everything on the other side.

$\displaystyle y \left( \frac{-3}{x-4} + \frac{4cos(x)}{sin(x)} -\frac{4x+4}{x^2+2x} + 3x^2 \right)$

we multiplied both sides by y, you can't just multiply one term on the right. what you see in brackets is what should go in the space provided

5. thanks, forgot to put in the brackets