JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

y = {e^{a{{\sin }^{ - 1}}x}} \Rightarrow \left( {1 - {x^2}} \right){y_{n + 2}} - \left( {2n + 1} \right)x{y_{n + 1}}

is equal to

View answer and explanation

Correct option: C

\left( {{n^2} + {a^2}} \right){y_n}

Option C (

\left( {{n^2} + {a^2}} \right){y_n}

) is correct because it is the correct result obtained by applying Leibnitz's rule to the given differential equation.

More Questions from DIFFERENTIATION

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x

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[x]

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\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}}