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

Let cos^ 1x=θ i.e., x=cos θ ∴ y=sec^ 11/2x^2 1⇒ = sec^ 1( 1/2cos^2θ 1)⇒ y = sec^ 1( 1/cos 2θ) ∵ cos 2θ= 2 cos^2θ 1 ⇒ y =sec^ 1(sec 2θ) ⇒ =2θ ⇒ y=2c ...

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Byju's Answer

Standard XII

Mathematics

Higher Order Derivatives

Find d y/d x ...

Question

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

$y = s e c^{- 1} (\frac{1}{2 x^{2} - 1}), 0 < x < \frac{1}{\sqrt{2}}$

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Solution

Let $c o s^{- 1} x = θ i . e ., x = c o s θ$

$∴ y = s e c^{- 1} \frac{1}{2 x^{2} - 1} \Rightarrow= s e c^{- 1} (\frac{1}{2 c o s^{2} θ - 1}) \Rightarrow y = s e c^{- 1} (\frac{1}{c o s 2 θ}) ∵ c o s 2 θ = 2 c o s^{2} θ - 1$

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

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

$\Rightarrow \frac{d y}{d x} = 2 \frac{d}{d x} (c o s^{- 1} x) = \frac{- 2}{\sqrt{1 - x^{2}}} (∵ \frac{d}{d x} (c o s^{- 1} x) = \frac{- 1}{\sqrt{1 - x^{2}}})$

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Find dydxin the following questions: y=sec−1(12x2−1),0<x<1√2

Let cos−1x=θ i.e.,x=cos θ ∴ y=sec−112x2−1⇒=sec−1(12cos2θ−1)⇒y=sec−1(1cos 2θ) ∵cos 2θ=2cos2θ−1 ⇒ y=sec−1(sec 2θ) ⇒ =2θ ⇒ y=2cos−1x) Differentiating both sides w.r.t. x, we get ⇒ dydx=2ddx(cos−1x)=−2√1−x2 (∵ ddx(cos−1x)=−1√1−x2)

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

$y = s e c^{- 1} (\frac{1}{2 x^{2} - 1}), 0 < x < \frac{1}{\sqrt{2}}$