# Calculus

Find y" by implicit differentiation.

x^3+y^3 = 1

(x^3)'+(y^3)' = (1)'
3x^2+3y^2(y') = 0
3y^2(y') = -3x^2
y' = -3x^2/3y^2
y' = -x^2/y^2

y" = [(y^2)(-x^2)'-(-x^2)(y^2)']/y^4
y" = [(y^2)(-2x)-(-x^2)(2y)]/y^4
y" = [(-2xy^2)-(-2yx^2)]/y^4
y" = -2xy(y-x)/y^4
y" = [-2x(y-x)]/y^3

My book shows the answer is y" = -2x/y^5.

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3. 👁 192
1. Your second line should be (via the chain rule!)

y" = [(y^2)(-2x)-(-x^2)(2yy')]/y^4
Now plug in -x^2/y^2 for y' and you should get the right result.

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posted by Steve
2. x^3+y^3 = 1

(x^3)'+(y^3)' = (1)'
3x^2+3y^2(y') = 0
3y^2(y') = -3x^2
y' = -3x^2/3y^2
y' = -x^2/y^2
GOOD

y" = [(y^2)(-x^2)'-(-x^2)(y^2)']/y^4
then I disagree with next line
y" = [(y^2)(-2x)-(-x^2)(2y DY/DX)]/y^4
WHERE dy/dx = -x^2/y^2

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posted by Damon
3. So chain rule is necessary since x and y are different variables on one side, forming a composition?

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posted by Justin
4. the chain rule is always necessary. It just happen s that dx/dx = 1, so it doesn't enter into the complexity.

x^3 + y^3 + u^2 = uv^2
Then you'd have

3x^2 + 3y^2 y' + 2u u' = u' v^2 = 2uv v'

Whenever you take a derivative, you have to factor in the chain rule.

It might be the case that x and y are both functions of t. Then you have

x^2 + y^2 = 10
2x dx/dt + 2y dy/dt = 0

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posted by Steve

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