Find d y-d x in the following questions-y -sin -11- x 2-1- x 2- 0- x -1

Substitute x = tan^ 1x=θ ∴ y=sin^ 1( 1 tan^2θ/1+tan^2θ) =sin^ 1(cos 2θ) ⇒ y=sin^ 1{sin ( π/x 2θ) } ⇒ y=π/2 2θ → y=π/x 2tant^ 1 x Differentiating both sid ...

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Standard XII

Mathematics

Basic Inverse Trigonometric Functions

Find d y/d x ...

Question

Find $\frac{d y}{d x}$ in the following questions:

$y = s i n^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), 0 < x < 1.$

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Solution

Substitute x = $t a n^{- 1} x = θ$

$∴ y = s i n^{- 1} (\frac{1 - t a n^{2} θ}{1 + t a n^{2} θ}) = s i n^{- 1} (c o s 2 θ) \Rightarrow y = s i n^{- 1} {s i n (\frac{π}{x} - 2 θ)}$

$\Rightarrow y = \frac{π}{2} - 2 θ \to y = \frac{π}{x} - 2 t a n t^{- 1} x$

Differentiating both sides w.r.t. x, we get

$\frac{d y}{d x} = \frac{d}{d x} (\frac{π}{2}) - 2 \frac{d}{d x} (t a n t^{- 1} x)$

$\frac{d y}{d x} = 0 - \frac{2}{1 + x^{2}} \Rightarrow \frac{d y}{d x} = \frac{- 2}{1 + x^{2}} (\frac{d}{d x} t a n^{- 1} x = \frac{1}{1 + x^{2}})$

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