JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

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Analyzing +4 | -1

{\cos ^{ - 1}}\left( {\frac{y}{2}} \right) = {\log _e}{\left( {\frac{x}{5}} \right)^5},\left| y \right| < 2,

then:

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Correct option: D

{x^2}y'' + xy' + 25y = 0

Option D (

{x^2}y'' + xy' + 25y = 0

) is correct because after differentiating the given equation twice and simplifying, it satisfies this relation.

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

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

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

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