JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

The derivative of

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

with respect to

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

, is

View answer and explanation

Correct option: C

1 for

|x|<1

and

-1

for

|x|>1

Option C (1 for |x|<1 and -1 for |x|>1) is correct because by using the substitution

x = tan(\theta)

, we can simplify both inverse trigonometric functions and then find the derivative.

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x

y

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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}}

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