If y-tan -1 x- then d y-d x at x-1 is equal to-

The correct option is C 1/√(2) Let y = sec (tan^ 1x) and tan^ 1 x = θ. ⇒ x = tan θ ⇒ y = √(1+x^2) (∵ sec^2 θ = 1 + tan^2 θ) ⇒dy/dx = 1/2√(1+x^2).2x At x = 1, ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Inverse Trigonometric Functions

If y=tan -1 x...

Question

If y = $s e c (t a n^{- 1} x)$ , then $\frac{d y}{d x}$ at x = 1 is equal to:

$\frac{1}{\sqrt{2}}$

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$\frac{1}{2}$

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$1$

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$0$

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Solution

The correct option is A
$\frac{1}{\sqrt{2}}$

Let $y = s e c (t a n^{- 1} x) a n d t a n^{- 1} x = θ . \Rightarrow x = t a n θ \Rightarrow y = \sqrt{1 + x^{2}} (∵ s e c^{2} θ = 1 + t a n^{2} θ) \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sqrt{1 + x^{2}}} .2 x A t x = 1, \frac{d y}{d x} = \frac{1}{\sqrt{2}} .$

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If y = sec(tan−1x), then dydx at x = 1 is equal to:

The correct option is A 1√2 Let y=sec(tan−1x) and tan−1x=θ.⇒x=tan θ⇒y=√1+x2(∵sec2θ=1+tan2θ)⇒dydx=12√1+x2.2xAt x=1,dydx=1√2.

If y = $s e c (t a n^{- 1} x)$ , then $\frac{d y}{d x}$ at x = 1 is equal to: