How to find dy/dx, if y=ln(8x^2+9y^2) ?

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Answer 1 · 2016-11-03T19:37:26

$dy/dx=(16x)/(8x^2+9y^2-18y)$

$y=ln(8x^2+9y^2)$

$dy/dx=(16x)/(8x^2+9y^2)+(18y)/ (8x^2+9y^2) dy/dx$

$dy/dx -(18y)/ (8x^2+9y^2) dy/dx=(16x)/(8x^2+9y^2)$

$dy/dx(1-(18y)/ (8x^2+9y^2) )=(16x)/(8x^2+9y^2)$

$dy/dx((8x^2+9y^2-18y)/ (8x^2+9y^2) )=(16x)/(8x^2+9y^2)$

$dy/dx=((16x)/(8x^2+9y^2))/((8x^2+9y^2-18y)/ (8x^2+9y^2) )$

$dy/dx=(16x)/(8x^2+9y^2-18y)$