JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left( {\frac{{a - x}}{{1 + ax}}} \right)} \right]

is equal to

View answer and explanation

Correct option: A

- \frac{1}{{1 + {x^2}}}

Option A (

\frac{{-1}}{{1 + x^2}}

) is correct because the expression inside the derivative simplifies to

\tan^{-1}(a) - \tan^{-1}(x)

, and the derivative of

\tan^{-1}(x)

\frac{1}{{1 + x^2}}

More Questions from DIFFERENTIATION

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x

y

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x y=1

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[x]

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x=k, k

f(x)

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

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

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

z = y + f\left( v \right),

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v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}