We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
65
2
avatar+301 

Find y'' by implicit differentiation.

 

5x^3 + 4y^3 = 6

 Mar 7, 2019
 #2
avatar+21978 
+2

Find y'' by implicit differentiation.

5x^3 + 4y^3 = 6

 

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

 

laugh

 Mar 7, 2019
edited by heureka  Mar 7, 2019

27 Online Users

avatar
avatar