derivative of 2xy + pi siny = 2pi
- Thread starter xc630
- Start date
pi cos y (dy/dx)?
what do i do with the pi in front
leaving it in front I got x + pi cos y (dy/dx) =-2y
dy/dx = -2y/ x+ pi cos y
plugging in the point (1, pi/2) I got -pi instead of -pi/2 ehich it should be. still i dont know where i went wrong
what do i do with the pi in front
leaving it in front I got x + pi cos y (dy/dx) =-2y
dy/dx = -2y/ x+ pi cos y
plugging in the point (1, pi/2) I got -pi instead of -pi/2 ehich it should be. still i dont know where i went wrong
skeeter
Elite Member
- Joined
- Dec 15, 2005
- Messages
- 3,204
your derivative is incorrect ...
\(\displaystyle \L 2xy + \pi \sin{y} = 2\pi\)
\(\displaystyle \L 2x \frac{dy}{dx} + 2y + \pi \cos{y} \frac{dy}{dx} = 0\)
\(\displaystyle \L 2x \frac{dy}{dx} + \pi \cos{y} \frac{dy}{dx} = -2y\)
\(\displaystyle \L \frac{dy}{dx}(2x + \pi \cos{y}) = -2y\)
\(\displaystyle \L \frac{dy}{dx} = \frac{-2y}{2x + \pi \cos{y}}\)
\(\displaystyle \L 2xy + \pi \sin{y} = 2\pi\)
\(\displaystyle \L 2x \frac{dy}{dx} + 2y + \pi \cos{y} \frac{dy}{dx} = 0\)
\(\displaystyle \L 2x \frac{dy}{dx} + \pi \cos{y} \frac{dy}{dx} = -2y\)
\(\displaystyle \L \frac{dy}{dx}(2x + \pi \cos{y}) = -2y\)
\(\displaystyle \L \frac{dy}{dx} = \frac{-2y}{2x + \pi \cos{y}}\)