Implicit Differentiation?

Question

When finding the second derivative implicitly, I know that you take the derivative of the first derivative.....but I'm not sure how to do that implicitly.

For example:

x^2+y^2=4
2x+2yy'=0
y'(2y)=-2x
y'=(-2x)/(2y)

Now what do I do? I do know that the answer is -4/(y^3)...

-4/(y^3) is the second derivative. I'm attempting to find the second derivative.

Answer 1

Begin with where you left of:

2x + 2yy' = 0

Differentiate again, just like the first time. Use the chain rule on the yy' term:

2 + 2(y'y' + yy'') = 0
1 + y'^2 + yy'' = 0
y'' = (-1-y'^2)/ y

From your y' = -x/y

y'' = (-1-(-x/y)^2)/ y = -x^2/y^3 - 1/y

This isn't the answer you have but see if they are the same:

x^2 = 4 - y^2

y'' = -(4-y^2)/y^3 - 1/y = -4/y^3 + 1/y - 1/y = -4/y^3

So the two solutions are the same

Answer 2

Use the Quotient Rule:

y'=(-2x)/(2y) = -x/y

Let u = -x and v = y

du/dx = -1 and dv/dx = dy/dx

y'' = (v*du/dx - u*dv/dx) / v^2

y'' = [-y - (-x)(dy/dx)] / (y^2)

since y' = (-x)/(y):

y'' = [-y + x * (-x/y)] / y^2)

y'' = [-y - (x^2/ y)] / y^2

Multiplying by y onto the numerator and denomninator, we have:

y'' = [ -y^2 - x^2] / y^3

But x^2 = 4 - y^2 from the equation:

y'' = [-y^2 - (4- y^2] / y^3

Therefore y'' = -4/ y^3

Answer 3

You can cancel the 2's
y' = (-2x)/(2y) = -x/y

I'm not sure where you got -4/y^3?

Answer 4

x^2 + y^2 = 4

2x + 2yy' = 0

2(x+yy') = 0

x + yy' = 0

yy'= -x

y'= -x/y

That is the final answer.

Answer 5

y' = (-2x)/(2y)
y' = -x/y
y" = (-y + xy')/y^2
y" = (-y + x(-x/y))/y^2
y" = (-y^2 - x^2)/y^3
y" = -(y^2 + x^2)/y^3
y" = -4/y^3