How do you find $\frac{d y}{d x}$ $dy/dx$ by implicit differentiation given $y^{2} = ln (2 x + 3 y)$ $y^2=ln(2x+3y)$ ?

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Answer 1 · 2017-02-21T14:53:52

$\frac{d y}{d x} = \frac{2}{6 y^{2} + 4 x y - 3} .$ $dy/dx=2/(6y^2+4xy-3).$

$y^{2} = ln (2 x + 3 y)$ $y^2=ln(2x+3y)$

$:. d/dx(y^2)=d/dx{ln(2x+3y)}.$

Using the Chain Rule, we have,

$d/dy(y^2)dy/dx=1/(2x+3y)d/dx(2x+3y).$

$:. 2ydy/dx=1/(2x+3y){2+3dy/dx}.$

$=2/(2x+3y)+3/(2x+3y)dy/dx.$

$:. {2y-3/(2x+3y)}dy/dx=2/(2x+3y).$

$:., if (2x+3y)ne0, {(4xy+6y^2-3)/cancel(2x+3y)}dy/dx=2/cancel(2x+3y).$

As, $(2x+3y)=0$ makes $ln(2x+3y)$ undefined, we have,

$:. dy/dx=2/(6y^2+4xy-3).$

How do you find dy/dxdydxdy/dx by implicit differentiation given y^2=ln(2x+3y)y2=ln(2x+3y)y^2=ln(2x+3y)?