1. ## Implicit differentiation

I'm having trouble getting:

ln(x/y)=(e^-x)(cos(y/2))

down to where I can separate y'. I was able to get it down to:

(1/x)-(1/y)(y')=e^-x (sin(y/2)(y'/2))

d/dx cos(y/2)= -sin(y/2)(y'/2) correct?

2. Originally Posted by ppark
ln(x/y)=(e^-x)(cos(y/2))

down to where I can separate y'. I was able to get it down to:

(1/x)-(1/y)(y')=e^-x (sin(y/2)(y'/2))

Check the right side:

$\displaystyle -e^{-x}\cos (y/2)+\ldots$

3. Originally Posted by ppark
I'm having trouble getting:

ln(x/y)=(e^-x)(cos(y/2))

down to where I can separate y'. I was able to get it down to:

(1/x)-(1/y)(y')=e^-x (sin(y/2)(y'/2))

d/dx cos(y/2)= -sin(y/2)(y'/2) correct?
The derivative of the left hand side is:

$\displaystyle \displaystyle \frac{d}{dx}\left( \ln(x/y)\right)=\frac{y}{x}\frac{d}{dx}\left( \frac{x}{y}\right)=\frac{y}{x}\left(\frac{1}{y}-\frac{x}{y^2}y' \right)= \ ...$

so that looks OK.

Now the derivative of the right hand side:

$\displaystyle \displaystyle \frac{d}{dx}\left( e^{-x}\cos(y/2)\right)=-e^{-x}\cos(y/2)-e^{-x}\sin(y/2)\frac{y'}{2}=\ ...$

CB

4. Originally Posted by CaptainBlack
The derivative of the left hand side is:

$\displaystyle \displaystyle \frac{d}{dx}\left( \ln(x/y)\right)=\frac{y}{x}\frac{d}{dx}\left( \frac{x}{y}\right)=\frac{y}{x}\left(\frac{1}{y}-\frac{x}{y^2}y' \right)= \ ...$

so that looks OK.

Now the derivative of the right hand side:

$\displaystyle \displaystyle \frac{d}{dx}\left( e^{-x}\cos(y/2)\right)=-e^{-x}\cos(y/2)-e^{-x}\sin(y/2)\frac{y'}{2}=\ ...$

CB
Wow thank you, can't believe I forgot to apply the product rule...

actually I got this the first time I tried the problem, but the problem with doing it this way is when I get all the y' on one side there doesn't seem a way to take out the y'.

((2/y')-(y'/y))=(((-e^-x)(cos(y/2)))-(e^-x)(sin(y/2)-(1/x))

((2y-(y'^2))/(y'(y)))=...