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How do you find #(dy)/(dx)# given #x^3+y^4=0#?

Answer 1

Isabella Ackerman

#dy/dx= -(3x^2)/(4y^3)#

#x^3+y^4=0#

Applying implicit differention and the power rule:

#3x^2+4y^3dy/dx =0#

#dy/dx= -(3x^2)/(4y^3)#

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Answer 2

Adam Adkins

To find (\frac{dy}{dx}) given (x^3+y^4=0), differentiate both sides of the equation implicitly with respect to (x), then solve for (\frac{dy}{dx}).

Differentiating both sides with respect to (x):

[\frac{d}{dx}(x^3) + \frac{d}{dx}(y^4) = 0]

[3x^2 + 4y^3 \frac{dy}{dx} = 0]

Now, solve for (\frac{dy}{dx}):

[\frac{dy}{dx} = -\frac{3x^2}{4y^3}]

This is the expression for (\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.