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How do you differentiate #x^(2/3)+y^(2/3)=4#?

Answer 1

In this case of implicit differentiation consider that #y# is function of #x# so you get: #2/3x^(2/3-1)+2/3y^(2/3-1)dy/dx=0# #dy/dx=-(x^(-1/3))/(y^(-1/3))=-root3(y/x)#

Where #y=(4-x^(2/3))^(3/2)# from your original expression.

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Answer 2

Elijah Abram

To differentiate the equation (x^{2/3} + y^{2/3} = 4) implicitly with respect to (x), you would follow these steps:

Differentiate each term of the equation with respect to (x).
Use the chain rule for differentiating terms involving (y) with respect to (x).
Solve for (\frac{{dy}}{{dx}}) to find the derivative.

Starting with the given equation (x^{2/3} + y^{2/3} = 4), the steps are as follows:

Differentiate each term: [ \frac{{d}}{{dx}}(x^{2/3}) + \frac{{d}}{{dx}}(y^{2/3}) = \frac{{d}}{{dx}}(4) ]
Apply the chain rule for terms involving (y): [ \frac{{2}}{{3}}x^{-1/3}\frac{{dx}}{{dx}} + \frac{{2}}{{3}}y^{-1/3}\frac{{dy}}{{dx}} = 0 ]
Solve for (\frac{{dy}}{{dx}}): [ \frac{{dy}}{{dx}} = -\frac{{x^{-1/3}}}{{y^{-1/3}}}]

This is the derivative of (y) with respect to (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.