If you are given the derivative of an implicit function as [tex] y' = \frac{y}{2y+x} [/tex] how would you find all points (x,y) such that the slope at those points is 1/2? Ok so I did: [tex] \frac{y}{2y+x} = \frac{1}{2} [/tex] and got x = 0. So if I substitute x = 0 back into the original equation I get [tex] (0, \sqrt{2}) [/tex]. Would this be the correct method to solve this question? Also show that the slope is never equaled to 0. So I did [tex] \frac{y}{2y+x} = 0 [/tex]. If y = 0, then the original equation wouldnt make sense. Is this a valid response?(adsbygoogle = window.adsbygoogle || []).push({});

Any help is appreciated

Thanks

**Physics Forums - The Fusion of Science and Community**

# Derivatives of implicit functions

Have something to add?

- Similar discussions for: Derivatives of implicit functions

Loading...

**Physics Forums - The Fusion of Science and Community**