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How do you implicitly differentiate #-y= x^3y^2-3x^2y^3-7xy^4 #?

Answer 1

Isabella Ackerman

A useful timesaver is to use #y'# instead of #dy/dx# when differentiating y ... here is how it works:

#-y= x^3y^2-3x^2y^3-7xy^4#

Now, using the chain and product rules ...

#-y'= 3x^2y^2+x^3(2yy')-6xy^3-3x^2(3y^2y')-7y^4-7x(4y^3y')#

Next combine the #y'# terms on the left side of the equation ...

#y'[-1-x^3(2y)+3x^2(3y^2)+7x(4y^3)]=3x^2y^2-6xy^3-7y^4#

Finally, simplify and solve for #y'#

#y'=(3x^2y^2-6xy^3-7y^4)/[9x^2(y^2)+28x(y^3)-x^3(2y)-1]#

hope that helped

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Answer 2

Charles Arocha

To implicitly differentiate -y = x^3y^2 - 3x^2y^3 - 7xy^4 with respect to x, apply the product rule and chain rule where necessary:

-1(dy/dx) = 3x^2y^2(dx/dx) + x^3(2y(dy/dx) + 2y^2(dx/dx)) - 3(2x)(xy^3)(dx/dx) - 3x^2(y^3)(dy/dx) - 7y^4(dx/dx)

Simplify and solve for 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.