- #1

- 76

- 0

^{2}+ y

^{2}= 36.

I know it is a circle with radius 6. Is there a better way to find the derivative then:

y

^{2}= 36 - x

^{2}

y = (36 - x

^{2})

^{1/2}

y = 1/2(36 - x

^{2})

^{-1/2}(-2x)

y

^{'}= -x/(36-x

^{2})

^{1/2}

- Thread starter sonofjohn
- Start date

- #1

- 76

- 0

I know it is a circle with radius 6. Is there a better way to find the derivative then:

y

y = (36 - x

y = 1/2(36 - x

y

- #2

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- 0

- #3

James R

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[tex]x^2 + y^2 = 36[/tex]

[tex]2x + 2y \frac{dy}{dx} = 0[/tex]

[tex]\frac{dy}{dx} = -\frac{x}{y} = -\frac{x}{\sqrt{36 - x^2}}[/tex]

Are you familiar with implicit differentiation?

- #4

- 737

- 0

Hence, we can differentiate both sides with respect to x without isolating y. Differentiating the left hand side, we get 2x + 2yy' using the chain rule. On the right hand side, differentiation 36, a constant function, just gives us 0.

We can then solve for y': 2x + 2yy' = 0 so y' = -x/y. Notice that we have y' in terms of both x and y(x) instead of just in x; this is a hallmark of implicit differentiation. If you solve for y in terms of x and plug it in, you find that

[tex]\frac{dy}{dx} = \frac{-x}{\pm\sqrt{36-x^2}}[/tex]

depending on whether y was positive or negative. This is a more complete version of the answer you found by solving for y first.

- #5

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- #6

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- #7

HallsofIvy

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2: You then wrote "find the derivative of x

3: Everyone here has assumed you really

- #8

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Sorry for the mix up. Yes I meant solve for dy/dx.

2: You then wrote "find the derivative of x^{2}+ y^{2}= 36" which also makes no sense. Youcandifferentiate (both sides of) an equation but you have to specify with respect to what variable.

3: Everyone here has assumed you reallymeant"find the derivative of y with respect to x, assuming that x^{2}+ y^{2}= 36".

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