I'm having a bit of trouble finding the second implicit derivative for the equation $y^5 = x^8$.
I have the first derivative, which is $8x^7 / 5y^4$, but I'm having trouble after that. I did the quotient rule and got $280x^6y^4 - 160x^7y^3 / (20y^3)^2$. I'm not quite sure what to do after this. Any help would be appreciated. Thanks!
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$\begingroup$In your quotient rule differentiation, there was an error. The derivative of $5y^4$ with respect to $x$ should be $20y^3\frac{dy}{dx}$. Then substitute the value of $\frac{dy}{dx}$ that you found in your first calculation, and, possibly, simplify.
I prefer to do things a little differently. After the first differentiation, we have $$5y^4\frac{dy}{dx}=8x^7.$$ Now differentiate again. We get $$5y^4\frac{d^2 y}{dx^2}+20y^3\left(\frac{dy}{dx}\right)^2=56x^6.$$ Substitute the value you found for $\frac{dy}{dx}$, and solve for $\frac{d^2 y}{dx^2}$.
$\endgroup$ 2 $\begingroup$$${ y }^{ 5 }-x^{ 8 }=0\\ \\ 5{ y }^{ 4 }\frac { dy }{ dx } -8{ x }^{ 7 }=0\\ \frac { dy }{ dx } =\frac { 8{ x }^{ 7 } }{ 5{ y }^{ 4 } } \\ \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\frac { d }{ dx } \left( \frac { 8{ x }^{ 7 } }{ 5{ y }^{ 4 } } \right) =\frac { 56{ x }^{ 6 }\cdot 5{ y }^{ 4 }-8{ x }^{ 7 }\cdot 20{ y }^{ 3 }\frac { dy }{ dx } }{ { \left( 5{ y }^{ 4 } \right) }^{ 2 } } =\frac { 280{ x }^{ 6 }{ y }^{ 4 }-160{ x }^{ 7 }{ y }^{ 3 }\frac { 8{ x }^{ 7 } }{ 5{ y }^{ 4 } } }{ 25{ y }^{ 8 } } \\ =\frac { 280{ x }^{ 6 }{ y }^{ 4 }-256{ x }^{ 14 }{ y }^{ -4 } }{ 25{ y }^{ 8 } } $$
$\endgroup$ $\begingroup$When you have powers, logarithmic differentiation makes life easier.
Consider $$F=y^5-x^8=0$$ Using the implicit function theorem $$F'_x=-8x^7 \qquad F'_y=5y^4 \implies \frac{dy}{dx}=-\frac{F'_x}{F'_y}=\frac{8x^7}{5y^4}$$ So $$y'=\frac{8x^7}{5y^4}\implies \log(y')=\log(\frac 85)+7\log(x)-4\log(y)$$ Differentiate again $$\frac{y''}{y'}=\frac 7 x-4\frac {y'}y$$ which makes $$y''=y'\left(\frac 7 x-4\frac {y'}y \right)$$ Replace and simplify to get the result in Battani's answer.
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