<s>[INST] What is $\frac{d^2y}{dx^2}$ of $(y^5) + (x^5) = 8$ when you solve using implicit differentiation? 
Find the Second derivative with respect to $x$ of :
$$ y^5 + x^5 = 8$$

When I solved it I got $\frac yx$ as the second derivative but I don’t think that I am right. Could someone explain it to me.
 [/INST] Starting with
$y^5 + x^5 = 8, \tag 1$
we differentiate once to obtain
$5y^4 y' + 5x^4 = 0, \tag 2$
from which we may isolate $y'$:
$y' = -\dfrac{x^4}{y^4}; \tag 3$
differentiating this expression using the quotient rule yields
$y'' = -\dfrac{4x^3y^4 - 4x^4y^3y'}{y^8} = -\dfrac{4x^3y - 4x^4y'}{y^5}; \tag 4$
substitute (3) into (4):
$y'' = -\dfrac{4x^3y - 4x^4(-x^4/y^4)}{y^5} = -\dfrac{4x^3y^5 + 4x^8}{y^9}; \tag 5$
we can now in fact eliminate $y$ from the numerator using (1) in the form
$y^5 = 8 - x^5, \tag 6$
that is
$y'' = -\dfrac{4x^3y^5 + 4x^8}{y^9}$
$= -\dfrac{4x^3(8 - x^5) + 4x^8}{y^9} = -\dfrac{32x^3 - 4x^8 + 4x^8}{y^9} = -\dfrac{32x^3}{y^9}; \tag 7$
it is possible to use (6) once again to reduce the power of $y$ in the denominator, viz.
$y'' = -\dfrac{32x^3}{y^9} = -\dfrac{32x^3}{(8 - x^5)y^4}; \tag 8$
an expression which may have some utility, though I am somewhat skeptical as to its genuine advantage over, say, (7).
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