CUET UG 2025 — Mathematics Arithmetic

Given $y = \sin^{-1}x$y=sin−1x

The derivative of inverse sine:

$$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$$dxdy​=1−x2​1​

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To find the second derivative, rewrite the first derivative as:

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

Applying the chain rule:

$$\frac{d^2y}{dx^2} = -\frac{1}{2}(1-x^2)^{-3/2} \times (-2x)$$dx2d2y​=−21​(1−x2)−3/2×(−2x)

$$\frac{d^2y}{dx^2} = \frac{x}{(1-x^2)^{3/2}}$$dx2d2y​=(1−x2)3/2x​

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Computing $(1-x^2)\frac{d^2y}{dx^2}$(1−x2)dx2d2y​:

$$(1-x^2)\frac{d^2y}{dx^2} = (1-x^2) \times \frac{x}{(1-x^2)^{3/2}}$$(1−x2)dx2d2y​=(1−x2)×(1−x2)3/2x​

$$= \frac{x(1-x^2)}{(1-x^2)^{3/2}}$$=(1−x2)3/2x(1−x2)​

$$= \frac{x}{(1-x^2)^{1/2}}$$=(1−x2)1/2x​

$$= \frac{x}{\sqrt{1-x^2}}$$=1−x2​x​

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Since $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$dxdy​=1−x2​1​:

$$\frac{x}{\sqrt{1-x^2}} = x \times \frac{1}{\sqrt{1-x^2}}$$1−x2​x​=x×1−x2​1​

$$= x\frac{dy}{dx}$$=xdxdy​

Therefore, $(1-x^2)\frac{d^2y}{dx^2} = x\frac{dy}{dx}$(1−x2)dx2d2y​=xdxdy​