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Derivatives of implicit functions 
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#1
Oct705, 03:01 PM

P: 1,237

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?
Any help is appreciated Thanks 


#2
Oct705, 03:39 PM

P: 788

Well, I wouldn't word it "the original equation wouldn't make sense". Can you word this more mathematically or demonstrate why cannot equal 0?



#3
Oct705, 03:44 PM

P: 666

[tex]\frac{y}{2y+x}=\frac{1}{2}\quad (1)[/tex] and come up with x=0, I don't see how you get the square root of 2. Try plugging x=0 into equation (1). The y variable will cancel out and you get an identity. What does that tell you about the possible values of y? Alex 


#4
Oct705, 03:46 PM

P: 788

Derivatives of implicit functions



#5
Oct705, 03:47 PM

HW Helper
P: 661




#6
Oct705, 03:50 PM

P: 666

Alex 


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