# implicit differentiation help

• Oct 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)
• Oct 11th 2009, 03:10 PM
artvandalay11
for the second one, factor $\displaystyle e^x$ out, and combine like terms to get your answer
• Oct 11th 2009, 03:12 PM
Jester
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 $\displaystyle x$ in red (above) I agree with you. For the second, factor and $\displaystyle e^x$ and you'll see an answer given.
• Oct 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 :[
• Oct 11th 2009, 05:13 PM
Jester
Differentiating gives

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

so

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

or

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

Originally Posted by Danny
Differentiating gives

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

so

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

or

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

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?