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How do you differentiate #y^2-y=3x^2y-6x+3#?

Answer 1

Emma Aldridge

#dy/dx=(6xy-6)/(2y-1-3x^2)#

differentiate all terms #color(blue)"implicitly with respect to x"#

Note that #3x^2y# has to be differentiated using the #color(blue)"product rule"#

#2y.dy/dx-1.dy/dx=(3x^2.dy/dx+y.(6x))-6+0#

#rArr2ydy/dx-dy/dx-3x^2dy/dx=6xy-6#

#rArrdy/dx(2y-1-3x^2)=6xy-6#

#rArrdy/dx=(6xy-6)/(2y-1-3x^2)#

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Answer 2

Paisley Johnson

To differentiate the equation (y^2 - y = 3x^2y - 6x + 3), we treat (y) as the variable and differentiate each term separately with respect to (x) using implicit differentiation. After differentiation, we rearrange the terms to solve for (\frac{dy}{dx}).

Starting with the given equation: (y^2 - y = 3x^2y - 6x + 3), we differentiate each term:

(\frac{d}{dx}(y^2) - \frac{d}{dx}(y) = \frac{d}{dx}(3x^2y) - \frac{d}{dx}(6x) + \frac{d}{dx}(3))

Using the chain rule and product rule where necessary, we get:

(2y\frac{dy}{dx} - \frac{dy}{dx} = 3x^2\frac{dy}{dx} + 6x\frac{dy}{dx} - 6)

Now, we isolate (\frac{dy}{dx}):

(\frac{dy}{dx}(2y - 1 - 3x^2 - 6x) = -6)

(\frac{dy}{dx}(2y - 3x^2 - 6x - 1) = -6)

Finally, solving for (\frac{dy}{dx}):

(\frac{dy}{dx} = \frac{-6}{2y - 3x^2 - 6x - 1})

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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.