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So here is the problem:

Calculate the second class derivative on $(1,1)$ of the equation $x^4+y^4=2$

I found this problem on my proffesor's notes. However it doesn't state whether a partial or a total derivative must be calculated. My guess would be the total.

So my approach would be:

1) Name a function $F(x,y)=0$

2) differentiate F to find $y'=\frac{dy}{dx}$

3) differentiate the result again

4) Solve $y'$ for $x=1$

5) Solve $y''$ for $x=1, y'(1)$

Is this approach at any point correct? I'm totally sure there is something that I'm missing.

(my approach results in $y''(1)=0$)

Chiranjeev_Kumar
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user253724
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  • See this answer in order to understand what's being asked. – Git Gud Jul 10 '15 at 17:52
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    Several solvers have come up with a friendly interpretation of this problem, and have proposed solutions accordingly. Nevertheless I must say that the formulation underlined in skin-tone is mathematically disgusting in every respect. – Christian Blatter Jul 10 '15 at 18:49

2 Answers2

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we have $x^4+y^4=2$ we assume the the derivative exists and we get $$4x^3+4y^3y'=0$$ divided by $4$ we get $$x^3+y^3y'=0$$ and the same thing again: $3x^2+3y^2(y')^2+y^3y''=0$ with $$y'=-\frac{x^3}{y^3}$$ you will get an equation for $$y''$$ allone.

Dr. Sonnhard Graubner
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$$x^4+y^4=2.$$

$$y^4=2-x^4$$ On differentiating with respect to $x$, we have

$$4y^3y'=-4x^3$$

$$y^3y'=-x^3$$ Again differentiating with respect to $x$, we have

$$3y^2(y')^2+y^3y''=-3x^2$$

Now Put your conditions and get your answer.

Chiranjeev_Kumar
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  • So your method can also be used for finding any higher class total derivative when given a function of the form F(x,y) when the variables are actually in the form x, y(x). Right? Same thing applied to functions of the form F(x,y(x),z(x),...) ? – user253724 Jul 11 '15 at 13:19