Implicit Differentiation: Finding d^2y/dx^2 at a Given Point

athamz
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Homework Statement


find the d^2y/dx^2 if y^3 + y = 2 cos x at the point (0,1)


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The Attempt at a Solution



my dy/dx = (-2 sin x)/(3y^2 + 1)

I don't know how to find d^2y/dx^2?
And when and how will I use the oint (0,1)?
 
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Differentiate both sides of the equation twice with respect to x, using chain rule on the left-hand size (differentiate with respect to y first than with respect to x).

ehild
 
oh.. so I have..

3y^2 * y' + y' = -2 sin x

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

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

Right?
 
It is right. Get the second differential. But differentiate both sides of the equation y'(3y^2 + 1) = -2 sin x instead of the last one. It will be an easier process.

ehild
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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