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

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

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Answer 1 · 2016-01-08T13:19:37

To differentiate implicitly, simply differentiate each term with respect to $x$ $x$ , then rearrange for $\frac{d y}{d x}$ $dy/dx$ .

Answer: $\frac{d y}{d x} = \frac{3 x^{2} y + 1}{8 y - x^{3} - 1}$ $dy/dx=(3x^2y+1)/(8y-x^3-1)$

Explanation:

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

I will take it piece by piece for the sake of clarity.

$\frac{d}{d x} (4 y^{2}) = 8 y \frac{d y}{d x}$ $d/(dx)(4y^2)=8ydy/dx$

This is from applying the chain rule.

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

This is from applying the product rule.

$\frac{d}{d x} (y) = \frac{d y}{d x}$ $d/(dx)(y)=dy/dx$

$\frac{d}{d x} (x) = 1$ $d/(dx)(x)=1$

Putting that all together we get:

$8 y \frac{d y}{d x} = x^{3} \frac{d y}{d x} + 3 x^{2} y + \frac{d y}{d x} + 1$ $8ydy/dx=x^3dy/dx+3x^2y+dy/dx+1$

Rearrange:

$\frac{d y}{d x} (8 y - x^{3} - 1) = 3 x^{2} y + 1$ $dy/dx(8y-x^3-1)=3x^2y+1$

$\frac{d y}{d x} = \frac{3 x^{2} y + 1}{8 y - x^{3} - 1}$ $dy/dx=(3x^2y+1)/(8y-x^3-1)$

Sometimes it is also possible to find $y$ $y$ in terms of $x$ $x$ in the original equation and sub back to get $\frac{d y}{d x}$ $dy/dx$ only in terms of $x$ $x$ .

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

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How do you implicitly differentiate 4y^2= x^3y+y-x 4y2=x3y+y−x4y^2= x^3y+y-x ?

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