+26393 Find y'' by implicit differentiation.
5x^3 + 4y^3 = 6
\(\begin{array}{|l|rcll|} \hline & \mathbf{5x^3 + 4y^3} & \mathbf{=}& \mathbf{6} \\\\ \hline y'=\ ? & 5\cdot 3x^{3-1} + 4\cdot 3y^{3-1}\cdot y' &=& 0 \\ & 15x^2+12y^2y' &=& 0 \\ & \mathbf{y'} &\mathbf{=}& \mathbf{-\dfrac{5x^2}{4y^2}} \\ \hline y''=\ ? & 15x^2+12y^2y' &=& 0 \\ & 15\cdot2 x^{2-1}+12\cdot\left( (y^2)'\cdot y' +y^2\cdot y'' \right) &=& 0 \\ & 30 x^{2-1}+12\cdot\left( 2\cdot y^{2-1}y'\cdot y' +y^2\cdot y'' \right) &=& 0 \\ & 30 x+12\cdot\left( 2\cdot y(y')^2 +y^2\cdot y'' \right) &=& 0 \quad |\quad : 6 \\ & 5x+ 2\cdot\left( 2\cdot y(y')^2 +y^2\cdot y'' \right) &=& 0 \\ & 5x+ 4y(y')^2 + 2y^2\cdot y'' &=& 0 \quad |\quad \mathbf{y'=-\dfrac{5x^2}{4y^2}}\\ & 5x+ 4y\left(-\dfrac{5x^2}{4y^2}\right)^2 + 2y^2\cdot y'' &=& 0 \\ & 5x+ \dfrac{4y\cdot 25x^4}{16y^4} + 2y^2\cdot y'' &=& 0 \\ & 5x+ \dfrac{25x^4}{4y^3} + 2y^2\cdot y'' &=& 0 \quad |\quad \cdot 4y^3 \\ & 20xy^3+ 25x^4 + 8y^5\cdot y'' &=& 0 \\ & 5x(\underbrace{4y^3+ 5x^3}_{=6}) + 8y^5\cdot y'' &=& 0 \\ & 5x\cdot 6 + 8y^5\cdot y'' &=& 0 \\ & 30x + 8y^5\cdot y'' &=& 0 \quad |\quad :2 \\ & 15x + 4y^5\cdot y'' &=& 0\\ & 4y^5\cdot y'' &=& -15x \quad |\quad :4y^5 \\ & \mathbf{y''} &\mathbf{=}& \mathbf{-\dfrac{15x}{4y^5}} \\ \hline \end{array}\)
