Implicit DIfferentiation for the curve y + cosy = x + 1

[Speak_fromtheheart]

I'm having trouble with a implicit differentiation problem with trig.

1. For the curve y + cosy = x + 1

a) Write the equation for the vertical tangent to the curve.

y+cosy = x+1
dy - siny * dy/dx = 1
dy/dx = 1/1-siny Right?

Then, set the bottom of that equal to zero.

1-siny = 0
-siny = -1
Now I'm stuck. What do you do next?

Thanks!

 

[pka]

Re: Implicit DIfferentiation

$\displaystyle \frac {dy} {dx} = \frac {1} {1 - \sin (y)}$ Right?

Yes that is correct.
The point you want is $\displaystyle ( \frac{\pi} {2} - 1, \frac{\pi} {2})$.