How to find the derivative of x2 + y2 = 36?

[sonofjohn]

How would I go about finding the derivative of x² + y² = 36.

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

y² = 36 - x²
y = (36 - x²)^1/2
y = 1/2(36 - x²)^-1/2(-2x)
y^' = -x/(36-x²)^1/2

 

[jeffreydk]

Usually with cases like this where it is inconvenient to differentiate explicitly you can use implicit differentiation.

 

[James R]

Another way is implicit differentiation:

x^2 + y^2 = 36

2x + 2y \frac{dy}{dx} = 0

\frac{dy}{dx} = -\frac{x}{y} = -\frac{x}{\sqrt{36 - x^2}}

Are you familiar with implicit differentiation?

 

[Tedjn]

You may use implicit differentiation. If f(x) = g(x), then f'(x) = g'(x). Therefore, we can have f(x) = x² + y(x)² (where I have written y explicitly as a function of x) and g(x) = 36.

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

\frac{dy}{dx} = \frac{-x}{\pm\sqrt{36-x^2}}​

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

 

[sonofjohn]

Implicit differentiation. I have learned it, but obviously need to make more use of it. Thank you kind sirs.

 

[sonofjohn]

On a second note, would it be better to leave the final answer in terms of x and y. Or should I solve for y and find y' in terms of x only. I prefer x/y but if the question asks for y' how should I answer?

 

[HallsofIvy]

1: You titled this "differentiation of a circle" which makes no sense. You cannot differentiate a geometric figure!

2: You then wrote "find the derivative of x² + y² = 36" which also makes no sense. You can differentiate (both sides of) an equation but you have to specify with respect to what variable.

3: Everyone here has assumed you really meant "find the derivative of y with respect to x, assuming that x²+ y²= 36".

 

[sonofjohn]

1: You titled this "differentiation of a circle" which makes no sense. You cannot differentiate a geometric figure!

2: You then wrote "find the derivative of x² + y² = 36" which also makes no sense. You can differentiate (both sides of) an equation but you have to specify with respect to what variable.

3: Everyone here has assumed you really meant "find the derivative of y with respect to x, assuming that x²+ y²= 36".

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