If y-tan-1-1-x-2-1-x- theny-1-1-4y-1-1y-1-0y-1- does not exist-

The correct option is A y'(1)=1/4Putting x=tan θ, we have √(1+x^2) 1/x=sec θ 1/tan θ=1 cos θ/sin θ=tanθ/2 y=θ/2=1/2tan^ 1x y'(x)=1/2(1+x^2)⇒ y'(1)=1/4 ...

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Standard XII

Mathematics

Differentiation Using Substitution

If y=tan^-1√(...

Question

If $y = t a n^{- 1} \frac{\sqrt{1 + x^{2}} - 1}{x}$ , then

y^{'} (1) = 1

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y^{'} (1) = \frac{1}{4}

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y^{'} (1) = 0

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y^{'} (1)

does not exist.

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Solution

The correct option is B $y^{'} (1) = \frac{1}{4}$
Putting $x = t a n θ$ , we have
$\frac{\sqrt{1 + x^{2}} - 1}{x} = \frac{s e c θ - 1}{t a n θ} = \frac{1 - c o s θ}{s i n θ} = t a n \frac{θ}{2}$
$y = \frac{θ}{2} = \frac{1}{2} t a n^{- 1} x$
$y^{'} (x) = \frac{1}{2 (1 + x^{2})} \Rightarrow y^{'} (1) = \frac{1}{4}$

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If y=tan−1√1+x2−1x, then

The correct option is B y′(1)=14Putting x=tan θ, we have √1+x2−1x=sec θ−1tan θ=1−cos θsin θ=tanθ2 y=θ2=12tan−1x y′(x)=12(1+x2)⇒y′(1)=14

If $y = t a n^{- 1} \frac{\sqrt{1 + x^{2}} - 1}{x}$ , then