JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

\sec \left( {\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}} \right) = {e^a},

then

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

is equal to

View answer and explanation

Correct option: B

\frac{y}{x}

Option B (y/x) is correct because implicit differentiation and simplification leads to dy/dx = y/x.

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x

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[x]

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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),

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