If y=[3]{x²+[3]{x²+[3]{x²+∞}}} then \;dy/dx is

Correct Answer is : 2x3y2−1\;{2x}{3y²-1}3y2−12x

Correct Answer is : 2x3y2−1\;{2x}{3y²-1}3y2−12x

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Differential Calculus

Question : 19 of 69

Marks: +1, -0

y=\sqrt[3]{x^{2}+\sqrt[3]{x^{2}+\sqrt[3]{x^{2}+\dots\infty}}}

\;\frac{dy}{dx}

Solution:

Concept:

The infinite nested cube root can be expressed as a functional equation

y = \sqrt[3]{x^2 + y}

Explanation:

Given

y = \sqrt[3]{x^{2}+\sqrt[3]{x^{2}+\sqrt[3]{x^{2}+\dots\infty}}}

We can write

y = \sqrt[3]{x^2 + y}

Cube both sides:

y^3 = x^2 + y

Thus

y^3 - y = x^2

Differentiate implicitly with respect to

x

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

Factor

\frac{dy}{dx}

\frac{dy}{dx}(3y^2 - 1) = 2x

Therefore

\frac{dy}{dx} = \frac{2x}{3y^2 - 1}

Answer:

\frac{dy}{dx} = \frac{2x}{3y^2 - 1}

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