If f(x)=sin−1⁡x−βα−β and g(x)=Tan−1⁡x−βα−x then show that f1(x)=g1(x)⋅(β<x<α)

Let f(x)=sin−1⁡x−βα−βf1(x)=11−x−βα−β⋅ddxx−βα−β∵ddxsin−1⁡x=11−x2=1α−xα−β1α−βd(x−β)dx=1α−xα−β1α−β12x−β∴f′(x)=12(α−x)(x−β).....(1)And g(x)=Tan−1⁡x−βα−x ddxtan−1⁡x=11+x2g1(x)=11+x−βα−x⋅ddxx−βα−x=α−xα−β⋅12x−βα−x⋅(α−x)(1)−(x−β)(−1)(α−x)2=α−x(α−x)2(α−β)x−β⋅α−β(α−x)2=12(α−x)(x−β).....(2)from (1) & (2) ∴f1(x)=g1(x)

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If $f (x) = \sin^{- 1} \sqrt{\frac{x - β}{α - β}} and g (x) = {Tan}^{- 1} \sqrt{\frac{x - β}{α - x}}$ then show that $f^{1} (x) = g^{1} (x) \cdot (β < x < α)$

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Detailed Solution

Let $f (x) = \sin^{- 1} \sqrt{\frac{x - β}{α - β}}$

$f^{1} (x) = \frac{1}{\sqrt{1 - \frac{x - β}{α - β}}} \cdot \frac{d}{d x} \sqrt{\frac{x - β}{α - β}} (∵ \frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}})$

$= \frac{1}{\frac{\sqrt{α - x}}{\sqrt{α - β}}} \frac{1}{\sqrt{α - β}} \frac{d (\sqrt{x - β})}{d x}$

$= \frac{1}{\frac{\sqrt{α - x}}{\sqrt{α - β}}} \frac{1}{\sqrt{α - β}} \frac{1}{2 \sqrt{x - β}}$
$∴ f^{'} (x) = \frac{1}{2 \sqrt{(α - x) (x - β)}}$ .....(1)