JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

y = {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right),

then

y'\left( 0 \right)

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because it is the value of the derivative of the given function at

x=0

after simplifying using trigonometric substitution.

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\frac{{dy}}{{dx}}

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