How do you differentiate $12 = y e^{x - y^{3}} + e^{x} - y$ $12=ye^(x-y^3)+e^x-y$ ?

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How do you differentiate $12 = y e^{x - y^{3}} + e^{x} - y$ $12=ye^(x-y^3)+e^x-y$ ?

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Answer 1 · 2017-06-01T23:29:23

$\frac{- e^{x} - y e^{x - y^{3}}}{e^{x - y^{3}} - 3 y^{3} e^{x - y^{3}} - 1} = \frac{d y}{d x}$ $(-e^x-ye^(x-y^3))/(e^(x-y^3)-3y^3e^(x-y^3)-1) = dy/dx$

Explanation:

Use implicit differentiation:

$\frac{d}{d x} 12 = \frac{d}{d x} (y e^{x - y^{3}} + e^{x} - y)$ $d/dx12 = d/dx(ye^(x-y^3)+e^x-y)$

$0 = \frac{d y}{d x} e^{x - y^{3}} + y e^{x - y^{3}} (1 - 3 y^{2} \cdot \frac{d y}{d x}) + e^{x} - \frac{d y}{d x}$ $0 = dy/dxe^(x-y^3) + ye^(x-y^3)(1-3y^2*dy/dx) + e^x-dy/dx$

$0 = \frac{d y}{d x} e^{x - y^{3}} + y e^{x - y^{3}} - 3 y^{3} e^{x - y^{3}} \frac{d y}{d x} + e^{x} - \frac{d y}{d x}$ $0 = dy/dxe^(x-y^3) + ye^(x-y^3) - 3y^3e^(x-y^3)dy/dx + e^x - dy/dx$

$- e^{x} - y e^{x - y^{3}} = \frac{d y}{d x} (e^{x - y^{3}} - 3 y^{3} e^{x - y^{3}} - 1)$ $-e^x-ye^(x-y^3) = dy/dx (e^(x-y^3)-3y^3e^(x-y^3)-1)$

$\frac{- e^{x} - y e^{x - y^{3}}}{e^{x - y^{3}} - 3 y^{3} e^{x - y^{3}} - 1} = \frac{d y}{d x}$ $(-e^x-ye^(x-y^3))/(e^(x-y^3)-3y^3e^(x-y^3)-1) = dy/dx$

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How do you differentiate $12 = y e^{x - y^{3}} + e^{x} - y$ $12=ye^(x-y^3)+e^x-y$ ?

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Explanation:

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