How do you implicitly differentiate $- 1 = x y^{2} + 2 x^{2} y - e^{y} csc (3 \frac{x}{y})$ $-1=xy^2+2x^2y-e^ycsc(3x/y)$ ?

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How do you implicitly differentiate $- 1 = x y^{2} + 2 x^{2} y - e^{y} csc (3 \frac{x}{y})$ $-1=xy^2+2x^2y-e^ycsc(3x/y)$ ?

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Answer 1 · 2016-06-19T07:36:27

$\frac{d y}{d x} = - \frac{\frac{3 e^{y}}{y} csc (\frac{3 x}{y}) cot (\frac{3 x}{y}) + y^{2} + 4 x y}{2 x y + 2 x^{2} - e^{y} csc (\frac{3 x}{y}) + \frac{3 x}{y^{2}} csc (\frac{3 x}{y}) cot (\frac{3 x}{y})}$ $(dy)/(dx)=-[(3e^y)/ycsc((3x)/y)cot((3x)/y)+y^2+4xy}/(2xy+2x^2-e^ycsc((3x)/y)+(3x)/y^2csc((3x)/y)cot((3x)/y))$

Explanation:

As $- 1 = x y^{2} + 2 x^{2} y - e^{y} csc (3 \frac{x}{y})$ $-1=xy^2+2x^2y-e^ycsc(3x/y)$ , taking differential of each side

$y^{2} + 2 x y \frac{d y}{d x} + 4 x y + 2 x^{2} \frac{d y}{d x} - (e^{y} \frac{d y}{d x} csc (\frac{3 x}{y}) - e^{y} csc (\frac{3 x}{y}) cot (\frac{3 x}{y}) (\frac{3}{y} - \frac{3 x}{y^{2}} \frac{d y}{d x})) = 0$ $y^2+2xy(dy)/(dx)+4xy+2x^2(dy)/(dx)-(e^y(dy)/(dx)csc((3x)/y)-e^ycsc((3x)/y)cot((3x)/y)(3/y-(3x)/y^2(dy)/(dx)))=0$

or $(2 x y + 2 x^{2}) \frac{d y}{d x} - \frac{d y}{d x} (e^{y} csc (\frac{3 x}{y}) - \frac{3 x}{y^{2}} csc (\frac{3 x}{y}) cot (\frac{3 x}{y})) + \frac{3 e^{y}}{y} csc (\frac{3 x}{y}) cot (\frac{3 x}{y}) = - y^{2} - 4 x y$ $(2xy+2x^2)(dy)/(dx)-(dy)/(dx)(e^ycsc((3x)/y)-(3x)/y^2csc((3x)/y)cot((3x)/y))+(3e^y)/ycsc((3x)/y)cot((3x)/y)=-y^2-4xy$

or $(2 x y + 2 x^{2} - e^{y} csc (\frac{3 x}{y}) + \frac{3 x}{y^{2}} csc (\frac{3 x}{y}) cot (\frac{3 x}{y})) \frac{d y}{d x} = - [\frac{3 e^{y}}{y} csc (\frac{3 x}{y}) cot (\frac{3 x}{y}) + y^{2} + 4 x y}$ $(2xy+2x^2-e^ycsc((3x)/y)+(3x)/y^2csc((3x)/y)cot((3x)/y))(dy)/(dx)=-[(3e^y)/ycsc((3x)/y)cot((3x)/y)+y^2+4xy}$

or $\frac{d y}{d x} = - \frac{\frac{3 e^{y}}{y} csc (\frac{3 x}{y}) cot (\frac{3 x}{y}) + y^{2} + 4 x y}{2 x y + 2 x^{2} - e^{y} csc (\frac{3 x}{y}) + \frac{3 x}{y^{2}} csc (\frac{3 x}{y}) cot (\frac{3 x}{y})}$ $(dy)/(dx)=-[(3e^y)/ycsc((3x)/y)cot((3x)/y)+y^2+4xy}/(2xy+2x^2-e^ycsc((3x)/y)+(3x)/y^2csc((3x)/y)cot((3x)/y))$

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How do you implicitly differentiate $- 1 = x y^{2} + 2 x^{2} y - e^{y} csc (3 \frac{x}{y})$ $-1=xy^2+2x^2y-e^ycsc(3x/y)$ ?

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How do you implicitly differentiate -1=xy^2+2x^2y-e^ycsc(3x/y) −1=xy2+2x2y−eycsc(3xy)-1=xy^2+2x^2y-e^ycsc(3x/y) ?

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How do you implicitly differentiate $- 1 = x y^{2} + 2 x^{2} y - e^{y} csc (3 \frac{x}{y})$ $-1=xy^2+2x^2y-e^ycsc(3x/y)$ ?