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What is the derivative of # (x^3)-(xy)+(y^3)=1#?

Answer 1

#dy/dx=(y-3x^2)/(3y^2-x)#

We need to find the derivative with respect to #x#, #dy//dx#.

When we differentiate #y#, since it's a function that's not #x#, the chain rule will kick in and a #dy//dx# term will arise thanks to the chain rule.

Also don't forget that differentiating #xy# will use the product rule.

Differentiating:

#d/dx(x^3-xy+y^3)=d/dx(1)#

#3x^2-(d/dxx)y-x(d/dxy)+3y^2(d/dxy)=0#

#3x^2-y-xdy/dx+3y^2dy/dx=0#

Solve for #dy/dx#:

#dy/dx(3y^2-x)=y-3x^2#

#dy/dx=(y-3x^2)/(3y^2-x)#

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Answer 2

Charles Anctil

To find the derivative of the implicit function (x^3 - xy + y^3 = 1), differentiate both sides of the equation with respect to (x), applying the chain rule where necessary.

[ \frac{{d}}{{dx}}(x^3 - xy + y^3) = \frac{{d}}{{dx}}(1) ]

[ 3x^2 - (x\frac{{dy}}{{dx}} + y) - (xy\frac{{dy}}{{dx}} + 3y^2\frac{{dy}}{{dx}}) = 0 ]

[ 3x^2 - x\frac{{dy}}{{dx}} - y - xy\frac{{dy}}{{dx}} - 3y^2\frac{{dy}}{{dx}} = 0 ]

[ 3x^2 - y - (x + 3y^2)\frac{{dy}}{{dx}} = 0 ]

[ (x + 3y^2)\frac{{dy}}{{dx}} = 3x^2 - y ]

[ \frac{{dy}}{{dx}} = \frac{{3x^2 - y}}{{x + 3y^2}} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.