High School Is dy/dx of x2+y2 = 50 the same as dy/dx of y = sqrt(50 - x2)?

[Blockade]

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)

 

[mathman]

Derivatives (dy/dx) are operators on functions, not on equations. Your post is meaningless.

 

[Mark44]

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.

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.

dy/dx = .5(-x²+50)^-.5*(-2x)

 

[WWGD]

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).