1
$\begingroup$

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 $y''(1)=0$)

$\endgroup$
  • $\begingroup$ See this answer in order to understand what's being asked. $\endgroup$ – Git Gud Jul 10 '15 at 17:52
  • 1
    $\begingroup$ 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. $\endgroup$ – Christian Blatter Jul 10 '15 at 18:49
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

$$x^4+y^4=2.$$

$$y^4=2-x^4$$ 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.

$\endgroup$
  • $\begingroup$ 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),...) ? $\endgroup$ – user253724 Jul 11 '15 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.