Implicit differentiation question

Question

Given that $x^n + y^n = 1$, show that $$\frac{d^2y}{dx^2} = -\frac{(n-1)x^{n-2}}{y^{2n-1}}.$$

I found that $\displaystyle nx^{n-1}+ny^{n-1}\frac{dy}{dx} = 0$ so that $\displaystyle y'=\frac{-x^{n-1}}{y^{n-1}}$.

Then $$n(n-1)x^{n-2}+n(n-1)y^{n-2}\left(\frac{dy}{dx}\right)^2 + \frac{d^2y}{dx^2}ny^{n-1} = 0.$$

Therefore $$y'' = \displaystyle \frac{-n(n-1)x^{n-2}-n(n-1)y^{n-2}(y')^2}{ny^{n-1}}.$$

Substituting the first derivative:

$$y'' = \displaystyle\frac{-n(n-1)x^{n-2}-n(n-1)y^{n-2}\left(\dfrac{-x^{n-1}}{y^{n-1}}\right)^2}{ny^{n-1}}.$$

I've been trying tons of different steps and can't establish which way to eliminate the $y$ terms from the numerator. Could someone offer a hint on how to proceed. Thanks.

Answer 1

Expanding your final expression, we have

\begin{align*} y'' &= \frac{-n(n-1)x^{n-2} -n(n-1)y^{n-2}\dfrac{x^{2n-2}}{y^{2n-2}}}{ny^{n-1}}\\ &= \frac{-n(n-1)x^{n-2} -n(n-1)y^{-n}x^{2n-2}}{ny^{n-1}}\\ &= \frac{-n(n-1)x^{n-2}(1 + x^{n}y^{-n})}{ny^{n-1}}\\ &= \frac{-(n-1)x^{n-2}(1 + x^{n}y^{-n})}{y^{n-1}}. \end{align*}

Now note that $x^n + y^n = 1$, so $x^ny^{-n} + 1 = y^{-n}$. Therefore

\begin{align*} y'' &= \frac{-(n-1)x^{n-2}(1 + x^{n}y^{-n})}{y^{n-1}}\\ &= \frac{-(n-1)x^{n-2}y^{-n}}{y^{n-1}}\\ &= \frac{-(n-1)x^{n-2}}{y^{2n-1}}. \end{align*}

Answer 2

Differentiating $y^{n-1}y_1+x^{n-1}=0$ wrt $x,$

$\displaystyle y^{n-1}y_2+(n-1)y^{n-2}y_1^2+(n-1)x^{n-2}=0$

$\implies \displaystyle -\frac{y^{n-1}y_2}{n-1}=x^{n-2}+y^{n-2}\left(-\frac{x^{n-1}}{y^{n-1}}\right)^2$ as $y_1=-\frac{x^{n-1}}{y^{n-1}}$

$$\implies -\frac{y^{n-1}y_2}{n-1}=x^{n-2}+\frac{x^{2n-2}}{y^n}=\frac{x^{n-2}(y^n+x^n)}{y^n}=\frac{x^{n-2}}{y^n}\text{ as } y^n+x^n=1$$

Answer 3

$y'=-x^{n-1}y^{1-n}$.

$y''=-(n-1)x^{n-2}y^{1-n}-x^{n-1}(1-n)y^{-n}y'$

$y''=-(n-1)x^{n-2}y^{1-n}+x^{n-1}(1-n)y^{-n}x^{n-1}y^{1-n}$

$y''=-(n-1)x^{n-2}y^ny^{1-2n}+x^{n-1}(1-n)x^{n-1}y^{1-2n}$

$y''=-(n-1)x^{n-2}(1-x^n)y^{1-2n}+x^{n-1}(1-n)x^{n-1}y^{1-2n}$

Should be easy from there.