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How do you implicitly differentiate #1=-3xy^2-xy#?

Answer 1

Christopher Allers

#y'=(-(3y^2+y))/((6xy+x)#

#D(-3xy^2-xy)=D(1)#

#:.-3[x.2yy'+y^2]-[xy'+y]=0#

#:.-6xy.y'-3y^2-xy'-y=0#

#:.-y'(6xy+x)=3y^2+y#

#:.y'=(-(3y^2+y))/((6xy+x))#

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Answer 2

William Abdalla

To implicitly differentiate (1=-3xy^2-xy), follow these steps:

Differentiate each term of the equation with respect to (x).
Apply the product rule and chain rule as necessary.
Solve for (\frac{{dy}}{{dx}}).

Differentiating each term:

[\frac{{d}}{{dx}}(1) = 0] [\frac{{d}}{{dx}}(-3xy^2) = -3y^2 \frac{{d}}{{dx}}(x) - 6xy \frac{{dy}}{{dx}}] [\frac{{d}}{{dx}}(-xy) = -y - x \frac{{dy}}{{dx}}]

Solve for (\frac{{dy}}{{dx}}):

[0 = -3y^2 - 6xy \frac{{dy}}{{dx}} - y - x \frac{{dy}}{{dx}}] [6xy \frac{{dy}}{{dx}} + x \frac{{dy}}{{dx}} = -3y^2 - y] [\frac{{dy}}{{dx}}(6xy + x) = -3y^2 - y] [\frac{{dy}}{{dx}} = \frac{{-3y^2 - y}}{{6xy + x}}]

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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.