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Hello,

I need to find the second derivative of 2x^2 - 3y^2 = 4 using implicit differentiation.

I got 2x/3y as my first derivative but am confused on how to solve for the second derivative using implicit differentiation.

 Jun 22, 2022
 #1
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This is what I would do...

 

dydx=2x3yd2ydx2=(3y)(2)(3dydx)(2x)(3y)2d2ydx2=(y)(2)(dydx)(2x)3(y)2d2ydx2=2y(2x3y)(2x)3y2d2ydx2=2y3y24x29y3d2ydx2=6y24x29y3

 

OR

 

2x23y2=44x6ydydx=046[y(d2ydx2)+(dydxdydx)=046[y(d2ydx2)+2x3y2x3y)=046[y(d2ydx2)+4x29y2]=046y(d2ydx2)64x29y2=03y(d2ydx2)=24x23y23y(d2ydx2)=6y24x23y2d2ydx2=6y24x29y3

 

 

 

 

 

LaTex

\frac{dy}{dx}=\frac{2x}{3y}\\
\frac{d^2y}{dx^2}=\frac{(3y)(2)-(3\frac{dy}{dx})(2x)}{(3y)^2}\\
\frac{d^2y}{dx^2}=\frac{(y)(2)-(\frac{dy}{dx})(2x)}{3(y)^2}\\
\frac{d^2y}{dx^2}=\frac{2y-(\frac{2x}{3y})(2x)}{3y^2}\\
\frac{d^2y}{dx^2}=\frac{2y}{3y^2}-\frac{4x^2}{9y^3}\\
\frac{d^2y}{dx^2}=\frac{6y^2-4x^2}{9y^3}\\

 Jun 27, 2022

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