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How do you differentiate #cos (y) -( x^2y^3) + 2y = pi#?

Answer 1

Emma Aldridge

This is a bit of a vague question, as it can be differentiated in terms of either #x# or #y# or both #x and y#

I will assume for #x and y#

#d^2/(dxdy) cos(y) - (x^2y^3) + 2y = d^2/(dxdy) pi#

we then first differentiate over #y# and treat #x# as a constant

#d/dx -sin(y) - (3x^2y^2) + 2 = d/dx 0#

then we can differentiate over #x# and treat #y# as a constant

# -1 - (6xy^2) = 0#

#6xy^2 = -1#

#xy^2 = -1/6#

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Answer 2

Paisley Johnson

To differentiate the given equation ( \cos(y) - x^2y^3 + 2y = \pi ) with respect to (x), we'll use implicit differentiation:

[ \frac{d}{dx}(\cos(y)) - \frac{d}{dx}(x^2y^3) + \frac{d}{dx}(2y) = \frac{d}{dx}(\pi) ]

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

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

Grouping terms involving ( \frac{dy}{dx} ), we get:

[ \left(-\sin(y) - 2xy^3 + 2\right)\frac{dy}{dx} = 3x^2y^2 ]

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

[ \frac{dy}{dx} = \frac{3x^2y^2}{-\sin(y) - 2xy^3 + 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.