How do you implicitly differentiate $- y = x y + 2 \sqrt{x - y^{3}}$ $-y=xy+2sqrt(x-y^3)$ ?

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How do you implicitly differentiate $- y = x y + 2 \sqrt{x - y^{3}}$ $-y=xy+2sqrt(x-y^3)$ ?

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Answer 1 · 2017-08-18T13:40:29

$\frac{d y}{d x} = - \frac{{(x - y^{3})}^{\frac{1}{2}}}{x {(x - y^{3})}^{\frac{1}{2}} + (1 - 3 y^{2}) + {(x - y^{3})}^{\frac{1}{2}}}$ $dy/dx=-(x-y^3)^(1/2)/(x(x-y^3)^(1/2)+(1-3y^2)+(x-y^3)^(1/2)$

Explanation:

$- \frac{d y}{d x}$ $-dy/dx$ $= x \frac{d y}{d x} + y +$ $=xdy/dx+y+$ ...

$2 \sqrt{x + y^{3}} =$ $2sqrt(x+y^3)=$ $2 {(x - y^{3})}^{\frac{1}{2}}$ $2(x-y^3)^(1/2)$ derivative of that equal to

${(x - y^{3})}^{- \frac{1}{2}} \cdot (1 - 3 y^{2} \frac{d y}{d x})$ $(x-y^3)^(-1/2)*(1-3y^2dy/dx)$

with some math

get $\frac{d y}{d x}$ $dy/dx$ in one side and all in the other side

$\frac{d y}{d x} = - \frac{{(x - y^{3})}^{\frac{1}{2}}}{x {(x - y^{3})}^{\frac{1}{2}} + (1 - 3 y^{2}) + {(x - y^{3})}^{\frac{1}{2}}}$ $dy/dx=-(x-y^3)^(1/2)/(x(x-y^3)^(1/2)+(1-3y^2)+(x-y^3)^(1/2)$

I need someone to check my answer or explain further, and I will be thankful.

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How do you implicitly differentiate $- y = x y + 2 \sqrt{x - y^{3}}$ $-y=xy+2sqrt(x-y^3)$ ?

1 Answer

Explanation:

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Science

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Humanities

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How do you implicitly differentiate -y=xy+2sqrt(x-y^3) −y=xy+2√x−y3-y=xy+2sqrt(x-y^3) ?

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Explanation:

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How do you implicitly differentiate $- y = x y + 2 \sqrt{x - y^{3}}$ $-y=xy+2sqrt(x-y^3)$ ?