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

Differentiating with respect to x :

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To implicitly differentiate the equation (x^{2/3} + y^{2/3} = 4), we apply the derivative to both sides of the equation with respect to (x), keeping in mind that (y) is a function of (x) (denoted as (y(x))). Therefore, when we differentiate terms involving (y), we use the chain rule.

Starting with the equation: [x^{2/3} + y^{2/3} = 4]

Differentiating both sides with respect to (x), we get: [\frac{d}{dx}(x^{2/3}) + \frac{d}{dx}(y^{2/3}) = \frac{d}{dx}(4)]

Applying the power rule and the chain rule, we obtain: [\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0]

Now, we solve for (\frac{dy}{dx}) to find the derivative of (y) with respect to (x): [\frac{2}{3}y^{-1/3}\frac{dy}{dx} = -\frac{2}{3}x^{-1/3}]

Divide both sides by (\frac{2}{3}y^{-1/3}) to isolate (\frac{dy}{dx}): [\frac{dy}{dx} = \left(-\frac{2}{3}x^{-1/3}\right) / \left(\frac{2}{3}y^{-1/3}\right)]

Simplifying, we have: [\frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}}]

Or equivalently: [\frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}}]

Therefore, the derivative of (y) with respect to (x), or the slope of the tangent to the curve at any point ((x, y)), is (-\frac{y^{1/3}}{x^{1/3}}).

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