# calculus4

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2y^3-y=7-x^4 find dy/dx using differentiating implicity

• calculus4 -

6y^2 dy/dx - dy/dx = 0 - 4x^3 dx/dx

notice I wrote down dx/dx even though it has a value of 1, and i wouldn't have to include it.
Since you are differentiating with respect to x, whenever you work on a purely x-term, the dx/dx does not have to be written.
BUT, whenever you are differentiating a variable other than x, d(?)/dx has to be included.
so..

6y^2 dy/dx - dy/dx = 0 - 4x^3 dx/dx
dy/dx(6y^2 - 1) = -4x^3
dy/dx = -4x^3/(6y^2 - 1)

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