Question

Suppose x^(8) e^(−3y) = ln (xy).

Find dy/dx by implicit differentiation.


Okay, is there a easy way of doing this or something? I got dy/dx = http://i.imgur.com/x7WJj.png
but I'm sure I made a mistake somewhere.

Can someone help me out?

Answer 1

This ones just ugly...\[x^8 e^{−3y} = \ln (xy)\]\[\frac {d}{dx} \left[ x^8 e^{−3y} = \ln (xy) \right]\]\[x^8 e^{−3y} = \ln (xy)\]\[8x^7 e^{-3y} - 3x^8 e^{-3y} y' = \frac {y + x y'}{xy}\]\[8x^8 e^{-3y}y - 3x^9 e^{-3y} y y' =y + x y'\]\[8x^8 e^{-3y}y - y = x y' + 3x^9 e^{-3y} y y'\]\[8x^8 e^{-3y}y - y = y' (x + 3x^9 e^{-3y} y )\]\[\frac {dy}{dx} = \frac {8x^8 e^{-3y}y - y}{x + 3x^9 e^{-3y} y}\]

Answer 2

Did I make any mistakes? Way too much stuff to keep track of :/

Answer 3

nope, completely right

Answer 4

\[x^8 e^{−3y} = \ln (xy)\]\[\frac {dy}{dx} = \frac {8x^8 e^{-3y}y - y}{x + 3x^9 e^{-3y} y} = \frac {y(8x^8 e^{-3y} - 1)}{x(1 + 3x^8 e^{-3y} y)}\]I think that can be simplified down to \[\frac {dy}{dx} = \frac {y(8 \ln xy - 1)}{x(1 + 3y lnxy)}\]

Answer 5

multiply by\[\frac{e^{3y}}{e^{3y}}\]and you get wolf's answer in case you don't believe me
http://www.wolframalpha.com/input/?i=implicit+differentiate+x%5E8+e%5E%28%E2%88%923y%29+%3D+ln+%28xy%29

Answer 6

That was some good practice :)

Answer 7

Okay I think I got it now thanks,