Trouble finding second implicit derivative

I have trouble finding the second implicit derivative.

This is the question.

Find y'' in terms of x and y by implicit differentiation.

$x^5 +y^5 = 2^5$

The final answer I always get is $\displaystyle -\frac{(4x^3)(32)}{y^9}$.

I might be doing something wrong

• Welcome to Math SE ! Could describe the first step ? Oct 17 '14 at 9:12
• As you can see from the answers, the problem is very simple once you obtained $y'$; for higher order derivatives $y^{(n)}$, you just need to express them as functions of lower orders. Oct 17 '14 at 9:44

First, $\displaystyle x^5+y^5=2^5 \rightarrow 5x^4+5y^4\cdot{y'}=0 \Rightarrow y'=-\frac{x^4}{y^4}$.

Second, $\displaystyle 5x^4+5y^4\cdot{y'}=0 \rightarrow 20x^3+20y^3\cdot{y'}+5y^4\cdot{y''}=0 \Rightarrow y''=-\frac{4x^3+4y^3\cdot{y'}}{y^4}$.

Using $\displaystyle y'=-\frac{x^4}{y^4}$ we will get $\displaystyle y''=-\frac{4x^3+4y^3\cdot{(-\frac{x^4}{y^4})}}{y^4}=-\frac{4x^3-\frac{4x^4}{y}}{y^4}=\frac{4x^3(x-y)}{y^5}$.

Take the derivative with respect to x, use y' and y'' to denote the first and second derivative of y with respect to x.
$\displaystyle 5x^4+5y^4y'=0$
Take the derivative again
$\displaystyle 20x^3+5(4y^3y'(y')+y^4y'')=0$
Rearrange to get,
$\displaystyle y''=\frac{-20x^3-20y^3(y')^2}{y^4}=\frac{-20x^3-20y^3(-x^4/y^4)^2}{y^4}$

• @frankooo, you should try using \displaystyle command in your LaTeX formulas. Oct 17 '14 at 9:09