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

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Answer 1 · 2016-01-08T13:19:37

To differentiate implicitly, simply differentiate each term with respect to $x$ , then rearrange for $dy/dx$ .

Answer: $dy/dx=(3x^2y+1)/(8y-x^3-1)$

$4y^2=x^3y+y-x$
$d/(dx)(4y^2)=d/(dx)(x^3y)+d/(dx)(y)-d/(dx)(x)$

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

$d/(dx)(4y^2)=8ydy/dx$

This is from applying the chain rule.

$d/(dx)(x^3y)=x^3dy/dx+3x^2y$

This is from applying the product rule.

$d/(dx)(y)=dy/dx$

$d/(dx)(x)=1$

Putting that all together we get:

$8ydy/dx=x^3dy/dx+3x^2y+dy/dx+1$

Rearrange:

$dy/dx(8y-x^3-1)=3x^2y+1$

$dy/dx=(3x^2y+1)/(8y-x^3-1)$

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

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