JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

y(\alpha)=\sqrt{2\left(\frac{\tan \alpha+\cot \alpha}{1+\tan ^{2} \alpha}\right)+\frac{1}{\sin ^{2} \alpha}}, \alpha \in\left(\frac{3 \pi}{4}, \pi\right)

, then

\frac{d y}{d \alpha}

\alpha=\frac{5 \pi}{6}

is:

View answer and explanation

Correct option: A

Option A (4) is correct because it is the value of the derivative of y with respect to alpha, evaluated at

\alpha = \frac{5\pi}{6}

More Questions from DIFFERENTIATION

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f(x)=|x|^{|\tan x|}

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x

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x y=1

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[x]

x

x=k, k

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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}}

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

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

v = \left( {\frac{x}{y}} \right),

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