implicit differentiation help

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October 11th 2009, 02:51 PM
tecknics

implicit differentiation help

I got a complete different answer from any of the choices, can anyone lead me to the right direction as to get an answer to this?

http://img188.imageshack.us/img188/3666/helpys.jpg

I ended up with xe^6x/cos(x+6y)-1/6, any help would be appreciated you thank :]

same as this one,
http://img27.imageshack.us/img27/5909/help1u.jpg

I ended up with e^x(x^2-2x+6)+e^x(2x-2)
October 11th 2009, 03:10 PM
artvandalay11

for the second one, factor $e^x$ out, and combine like terms to get your answer
October 11th 2009, 03:12 PM
Danny

Quote:

Originally Posted by tecknics

I got a complete different answer from any of the choices, can anyone lead me to the right direction as to get an answer to this?

http://img188.imageshack.us/img188/3666/helpys.jpg

I ended up with xe^6x/cos(x+6y)-1/6, any help would be appreciated you thank :]

same as this one,
http://img27.imageshack.us/img27/5909/help1u.jpg

I ended up with e^x(x^2-2x+6)+e^x(2x-2)

Outside of the $x$ in red (above) I agree with you. For the second, factor and $e^x$ and you'll see an answer given.
October 11th 2009, 04:49 PM
tecknics

i got the second one, rather easy dunno how i missed it, but as far as the first one goes im still lost :[
October 11th 2009, 05:13 PM
Danny

Differentiating gives

$ 6 e^{6x} = \cos(x+6y)(1 + 6y') $

so

$ 1+6y' = \frac{6e^{6x}}{\cos(x+6y)} $

or

$ y' = \frac{e^{6x}}{\cos(x+6y)} - \frac{1}{6}. $
October 11th 2009, 05:28 PM
tecknics

Quote:

Originally Posted by Danny

Differentiating gives

$ 6 e^{6x} = \cos(x+6y)(1 + 6y') $

so

$ 1+6y' = \frac{6e^{6x}}{\cos(x+6y)} $

or

$ y' = \frac{e^{6x}}{\cos(x+6y)} - \frac{1}{6}. $

Yes, i got that answer excluding my previous x, but its not an option on one of the given responses :( could the responses given to me be incorrect?