How do you differentiate $x^{4} (x + y) = y^{2} (3 x - y)$ $x^4(x+y)=y^2(3x-y)$ ?

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How do you differentiate $x^{4} (x + y) = y^{2} (3 x - y)$ $x^4(x+y)=y^2(3x-y)$ ?

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Answer 1 · 2015-09-11T06:54:35

$\frac{d y}{d x} = \frac{3 y^{2} - 5 x^{4} - 4 x^{3} y}{x^{4} - 6 x y + 3 y^{2}}$ $dy/dx=(3y^2-5x^4-4x^3y)/(x^4 -6xy+3y^2)$

Explanation:

This can be by implicit differentiation. Rewrite it as $x^{5} + x^{4} y = 3 x y^{2} - y^{3}$ $x^5+x^4 y= 3xy^2 - y^3$ . Now differentiate term wise, $5 x^{4} + x^{4} \frac{d y}{d x} + 4 x^{3} y = 3 y^{2} + 6 x y \frac{d y}{d x} - 3 y^{2} \frac{d y}{d x}$ $5x^4 +x^4 dy/dx + 4x^3 y= 3y^2+6xydy/dx-3y^2 dy/dx$

$(x^{4} - 6 x y + 3 y^{2}) \frac{d y}{d x} = 3 y^{2} - 5 x^{4} - 4 x^{3} y$ $(x^4 -6xy+3y^2)dy/dx=3y^2-5x^4-4x^3y$

$\frac{d y}{d x} = \frac{3 y^{2} - 5 x^{4} - 4 x^{3} y}{x^{4} - 6 x y + 3 y^{2}}$ $dy/dx=(3y^2-5x^4-4x^3y)/(x^4 -6xy+3y^2)$

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How do you differentiate $x^{4} (x + y) = y^{2} (3 x - y)$ $x^4(x+y)=y^2(3x-y)$ ?

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Explanation:

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How do you differentiate x^4(x+y)=y^2(3x-y)x4(x+y)=y2(3x−y)x^4(x+y)=y^2(3x-y)?

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How do you differentiate $x^{4} (x + y) = y^{2} (3 x - y)$ $x^4(x+y)=y^2(3x-y)$ ?