How do you implicitly differentiate #2= xy-ysin^2x-cos^2xy^2 #?

Answer 1

Use Leibniz notation and you should be fine. For the second and third terms, you have to apply chain rule a couple of times.

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Answer 2

To implicitly differentiate ( 2 = xy - y\sin^2(x) - \cos^2(xy^2) ), we differentiate each term with respect to ( x ) and then solve for ( \frac{dy}{dx} ).

[ \frac{d}{dx} (2) = \frac{d}{dx} (xy) - \frac{d}{dx} (y\sin^2(x)) - \frac{d}{dx} (\cos^2(xy^2)) ]

Simplify and differentiate each term:

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

Rearrange terms and solve for ( \frac{dy}{dx} ):

[ \frac{dy}{dx} (x - 2\cos(xy^2)y^2) = y\sin^2(x) - y ]

Finally,

[ \frac{dy}{dx} = \frac{y\sin^2(x) - y}{x - 2\cos(xy^2)y^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.

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.

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.

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.

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