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

Let tan^ 1x=θ i.e., x=tant θ ∴ y=cos^ 1(1 x^2/1+x^2) =y=cos^ 1(1 tan^2θ/1+tan^2θ) (∵ 1 tan^2θ/1+tan^2θ=cos 2θ) ⇒ y=cos^ 1(cos 2θ)=2 tan^ 1x Differentiat ...

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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:

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

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Solution

Let $t a n^{- 1} x = θ i . e ., x = t a n t θ$

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

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

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

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

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