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How do you implicitly differentiate # y-(3y-x)^2-1/y^2=x^3 + y^3- xy#?

Answer 1

Scarlett Allison

By differentiating #x# and #y# like your usual differetiation practices.

#y-(3y-x)^2-1/y^2=x^3+y^3-xy#

I will let #(dy)/(dx)=y'#

Now begin the implicit differentiation

#y'-2(3y-x)(3y'-1)-(-2)1/y^3y'=3x^2+3y^2y'-(y+xy')#

I moved all the terms to one side and factored out #y'#

#y'[7-18y+2/y^2-3y^2+x]+7y-2x-3x^2=0#

Trick is; after differentiating any form of #y#, add #y'#

Example: #d/dxy^2=2yy'#

Hope this helps. Cheers

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Answer 2

Aurora Alvarado

To implicitly differentiate the given equation, follow these steps:

Differentiate each term of the equation with respect to (x).
Apply the chain rule whenever necessary.
Simplify the resulting expression.

After differentiating, you'll obtain the derivative of (y) with respect to (x), denoted as (\frac{{dy}}{{dx}}).

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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.