Question #9d02e

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Answer 1 · 2017-11-03T08:12:53

$[-4y^2-x^2]/(16y^3)$

First, we can find $dy/dx$ by differentiating both sides.

$x^2+4y^2=4$

$d/dx(x^2)+d/dx(4y^2)=d/dx(4)$

$2x+(4*2y)*dy/dx=0$

$8y*dy/dx=-2x$

$dy/dx=(-2x)/(8y)$

$dy/dx=(-x)/(4y)$

Then, we can differentiate it again to get the second derivative.

$d/dx (dy/dx)=d/dx((-x)/(4y))$

$=[4y*d/dx(-x)-(-x)*d/dx(4y)]/(4y)^2$

$=[4y*(-1)-(-x)*4dy/dx]/(16y^2)$

$=[-4y+4xdy/dx]/(16y^2)$

Then, we can sub $dy/dx$ that we found in the above and get the final answer.

$=[-4y+4x((-x)/(4y))]/(16y^2)$

$=[-4y-x^2/y]/(16y^2)$

$=[-4y^2-x^2]/(16y^3)$

Here is the answer. Hope this can help you :)