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Q.If y=tan⁡xy = \tan\sqrt{x} then dydx=?\dfrac{dy}{dx} = ?

(a) sec⁡x2x\dfrac{\sec\sqrt{x}}{2\sqrt{x}}
(b) sec⁡x2x\dfrac{\sec x}{2\sqrt{x}}
(c) sec⁡2x2x\dfrac{\sec^2\sqrt{x}}{2\sqrt{x}}
(d) sec⁡2xx\dfrac{\sec^2\sqrt{x}}{\sqrt{x}}
Jharkhand JacJAC Intermediate Board (1st Year) 2025MCQ· 1mImportance★★★★★
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Differentiating y=tan⁡xy=\tan\sqrt x by the chain rule: ddxtan⁡u=sec⁡2u⋅dudx\frac{d}{dx}\tan u = \sec^2u\cdot\frac{du}{dx} with u=xu=\sqrt x, giving sec⁡2x2x\dfrac{\sec^2\sqrt x}{2\sqrt x}.

Let u=xu = \sqrt{x}, so y=tan⁡uy = \tan u.

dydu=sec⁡2u=sec⁡2x\frac{dy}{du} = \sec^2 u = \sec^2\sqrt{x}

dudx=12x\frac{du}{dx} = \frac{1}{2\sqrt{x}}

By the chain rule: …

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