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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Mathematics
Chapter
Differentiation

BITSAT 2024 · Mathematics · Differentiation

If y= ^ -1 ( x -x 1+x^ (3)/(2) ), then y^ (1) is equal to

If y=tan⁡−1(x−x1+x32),theny=\tan ^{-1} \left(\frac{\sqrt{x}-x}{1+x^{\frac{3}{2}}}\right), then y^{\prime}(1)isequaltois equal to

Options

  1. A0
  2. B12\frac{1}{2}
  3. C-1
  4. D−14-\frac{1}{4}

Correct answer

D. −14-\frac{1}{4}

Step-by-step solution

Given, y=tan⁡−1(x−x1+x32)y=tan⁡−1(x−x1+x⋅x)y=tan⁡−1(x)−tan⁡−1x[∵tan⁡−1(x−y1+xy)=tan⁡−1x−tan⁡−1y]\begin{aligned} y & =\tan ^{-1}\left(\frac{\sqrt{x}-x}{1+x^{\frac{3}{2}}}\right) \\ y & =\tan ^{-1}\left(\frac{\sqrt{x}-x}{1+\sqrt{x} \cdot x}\right) \\ y & =\tan ^{-1}(\sqrt{x})-\tan ^{-1} x \\ & {\left[\because \tan ^{-1}\left(\frac{x-y}{1+x y}\right)=\tan ^{-1} x-\tan ^{-1} y\right] }\end{aligned} Differentiating w.r.t xy′(x)=dydx=11+x×12x−11+x2xy^{\prime}(x)=\frac{d y}{d x}=\frac{1}{1+x} \times \frac{1}{2 \sqrt{x}}-\frac{1}{1+x^2} Now, y′(1)=12×12−12=−14\quad y^{\prime}(1)=\frac{1}{2} \times \frac{1}{2}-\frac{1}{2}=-\frac{1}{4}
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