Implicit Differentiation

  • #1
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For implicit differentiation, is dy/dx of x2+y2 = 50 the same as y2 = 50 - x2 ?

From what I can take it, it'd be a no since.
For x2+y2 = 50,
d/dx (x2+y2) = d/dx (50) --- will eventually be ---> dy/dx = -x/y

Where,
y2 = 50 - x2
y = sqrt(50 - x2)
dy/dx = .5(-x2+50)-.5*(-2x)
 

Answers and Replies

  • #2
mathman
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Derivatives (dy/dx) are operators on functions, not on equations. Your post is meaningless.
 
  • #3
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For implicit differentiation, is dy/dx of x2+y2 = 50 the same as y2 = 50 - x2 ?
You don't "take dy/dx of" anything. dy/dx is the derivative of y with respect to x. The two equations above are equivalent, meaning that any (x, y) pair that satisifies one equation also satisfies the other equation.

If you differentiate both sides of either equation with respect to x, you should be able to find dy/dx.
Blockade said:
From what I can take it, it'd be a no since.
For x2+y2 = 50,
d/dx (x2+y2) = d/dx (50) --- will eventually be ---> dy/dx = -x/y

Where,
y2 = 50 - x2
y = sqrt(50 - x2)
There's no need to solve for y. Just differentiate y2 implicitly with respect to x.
Blockade said:
dy/dx = .5(-x2+50)-.5*(-2x)
 
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  • #4
WWGD
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For implicit differentiation, is dy/dx of x2+y2 = 50 the same as y2 = 50 - x2 ?

From what I can take it, it'd be a no since.
For x2+y2 = 50,
d/dx (x2+y2) = d/dx (50) --- will eventually be ---> dy/dx = -x/y

Where,
y2 = 50 - x2
y = sqrt(50 - x2)
dy/dx = .5(-x2+50)-.5*(-2x)
Sub -in the value of y in the first equation ( and cancel out the twos on the second one).
 

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