New chat PYQ Library Doubts Rank report

Sign in to see Recents

Sign in to save progress →

Differentiation question

2021 · 27 Aug · Shift 2 · Q33

Guest · try one question or challengeGuest · try one question or challenge

PYQ Library
JEE Main
Mathematics
Differentiation
2021 · 27 Aug · Shift 2 · Q33

Differentiation question

2021 · 27 Aug · Shift 2 · Q33

JEE MainMathematicsDifferentiationMCQ+4 / −1

If

y(x) = {\cot ^{ - 1}}\left( {{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \over {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right),x \in \left( {{\pi \over 2},\pi } \right)

, then

{{dy} \over {dx}}

at

x = {{5\pi } \over 6}

is :

A
$- {1 \over 2}$
B
$-$ 1
C
${1 \over 2}$
D
0

More from Differentiation

If y = $k = 1 \sum 6 k cos^{- 1} {\frac{3}{5} cos k x - \frac{4}{5} sin k x}$ , then $\frac{d y}{d x}$ at x = 0 is .2020 · Numerical
If y2 + loge (cos2x) = y, $x \in (- \frac{π}{2}, \frac{π}{2})$ , then :2020 · MCQ
If $(a + 2 b cos x) (a - 2 b cos y) = a^{2} - b^{2}$ where a > b > 0, then $\frac{d x}{d y} a t (\frac{π}{4}, \frac{π}{4})$ is :2020 · MCQ
The derivative of $tan^{- 1} (\frac{1 + x ^{2} - 1}{x})$ with respect to $tan^{- 1} (\frac{2 x 1 - x ^{2}}{1 - 2 x ^{2}})$ at x = $\frac{1}{2}$ is :2020 · MCQ
Let xk + yk = ak, (a, k > 0 ) and $\frac{d y}{d x} + (\frac{y}{x})^{\frac{1}{3}} = 0$ , then k is:2020 · MCQ
If $y (α) = 2 (\frac{t a n α + c o t α}{1 + t a n ^{2} α}) + \frac{1}{s i n ^{2} α}, α \in (\frac{3 π}{4}, π) \frac{d y}{d α} a t α = \frac{5 π}{6} i s$ …2020 · MCQ
Let y = y(x) be a function of x satisfying $y 1 - x^{2} = k - x 1 - y^{2}$ where k is a constant and $y (\frac{1}{2}) = - \frac{1}{4}$ . Then $\frac{d y}{d x}$ at x = $\frac{1}{2}$ , is equal to :2020 · MCQ
Let ƒ(x) = (sin(tan–1x) + sin(cot–1x))2 – 1, |x| > 1. If $\frac{d y}{d x} = \frac{1}{2} \frac{d}{d x} (sin^{- 1} (f (x)))$ and $y (3) = \frac{π}{6}$ , then y( $- 3$ …2020 · MCQ