How do you implicitly differentiate $3 y + \frac{y^{4}}{x^{2}} = 2$ $3y + y^4/x^2 = 2$ ?

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How do you implicitly differentiate $3 y + \frac{y^{4}}{x^{2}} = 2$ $3y + y^4/x^2 = 2$ ?

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Answer 1 · 2016-12-08T19:23:59

For problems like these, we must remember that we're differentiating with respect to x.

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

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

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

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

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

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

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

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

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

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How do you implicitly differentiate $3 y + \frac{y^{4}}{x^{2}} = 2$ $3y + y^4/x^2 = 2$ ?

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