CUET UG 2025 — Mathematics Calculus

Given: $y = \frac{1}{\sqrt[3]{1-x^3}}$y=31−x3​1​

Rewriting in a simpler form for differentiation:

$$y = \frac{1}{(1-x^3)^{1/3}}$$y=(1−x3)1/31​

$$y = (1-x^3)^{-1/3}$$y=(1−x3)−1/3

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Applying the chain rule with power $n = -\frac{1}{3}$n=−31​ and inner function $f(x) = 1-x^3$f(x)=1−x3:

$$\frac{dy}{dx} = -\frac{1}{3} \cdot (1-x^3)^{-\frac{1}{3}-1} \cdot \frac{d}{dx}(1-x^3)$$dxdy​=−31​⋅(1−x3)−31​−1⋅dxd​(1−x3)

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Simplifying the exponent:

$$-\frac{1}{3} - 1 = -\frac{4}{3}$$−31​−1=−34​

$$\frac{dy}{dx} = -\frac{1}{3} \cdot (1-x^3)^{-4/3} \cdot \frac{d}{dx}(1-x^3)$$dxdy​=−31​⋅(1−x3)−4/3⋅dxd​(1−x3)

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The derivative of the inner function:

$$\frac{d}{dx}(1-x^3) = -3x^2$$dxd​(1−x3)=−3x2

Substituting:

$$\frac{dy}{dx} = -\frac{1}{3} \cdot (1-x^3)^{-4/3} \cdot (-3x^2)$$dxdy​=−31​⋅(1−x3)−4/3⋅(−3x2)

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Simplifying the constants:

$$\frac{dy}{dx} = -\frac{1}{3} \times (-3x^2) \times (1-x^3)^{-4/3}$$dxdy​=−31​×(−3x2)×(1−x3)−4/3

$$\frac{dy}{dx} = x^2(1-x^3)^{-4/3}$$dxdy​=x2(1−x3)−4/3

Therefore, $\frac{dy}{dx} = x^2(1-x^3)^{-4/3}$dxdy​=x2(1−x3)−4/3