- #1

- 241

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y'=(-2x-8x^3+2x^5)/(1+x^4)^2

y''=[ (1+x^4)^2(-2-24x^2+10x^4)-(-2x-8x^3+2x^5)8x^3(1+x^4) ] / (1+x^4)^4

=[ -2(1+x^4)(1+12x^2-5x^4)+16x^4(1+4x^2-x^4) ] /(1+x^4)^3

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- Thread starter nameVoid
- Start date

- #1

- 241

- 0

y'=(-2x-8x^3+2x^5)/(1+x^4)^2

y''=[ (1+x^4)^2(-2-24x^2+10x^4)-(-2x-8x^3+2x^5)8x^3(1+x^4) ] / (1+x^4)^4

=[ -2(1+x^4)(1+12x^2-5x^4)+16x^4(1+4x^2-x^4) ] /(1+x^4)^3

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- #2

Homework Helper

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noticing that the denominator is always > 0 and both the numerator & denominator are symmetric should also help

- #3

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i need to calculate the exact points of inflections

- #4

Homework Helper

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hmmm... how about considering a substitution u = x^2 then

this should hopefully lead to a simpler calculation of the zeroes of the derivatives w.r.t. x

then if ' denotes dervative w.r.t. x, using the chain rule

[tex] y' = \frac{dy}{dx} = \frac{d}{dx} y(u(x)) = \frac{d y(u)}{du} \frac{du(x)}{dx} = \frac{dy}{du} u'[/tex]

and so on for the next one, where you'll need the product rule as well

this should hopefully lead to a simpler calculation of the zeroes of the derivatives w.r.t. x

then if ' denotes dervative w.r.t. x, using the chain rule

[tex] y' = \frac{dy}{dx} = \frac{d}{dx} y(u(x)) = \frac{d y(u)}{du} \frac{du(x)}{dx} = \frac{dy}{du} u'[/tex]

and so on for the next one, where you'll need the product rule as well

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