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NDA Mathematics Differentiation 2025 NDA 2025 (Phase 1)

NDA Mathematics Question (2025) — Solution

Question

Consider the following for the two (02) items that follow: Let x = - and y = ^4 - ^4 . What is ( x^2+4 y^2+4 ) dy dx [(x^2+4) d^2y dx^2 - 16y ] equal to?

Options

  1. A. 16x
  2. B. 16y
  3. C. -16x
  4. D. -16y

Answer

C. -16x

Step-by-step solution

Given x = - and y = ^4 - ^4 . First, we simplify x^2 + 4 and y^2 + 4: x^2 + 4 = ( - )^2 + 4 = ^2 + ^2 - 2 + 4 = ( + )^2 y^2 + 4 = ( ^4 - ^4 )^2 + 4 = ^8 + ^8 - 2 + 4 = ( ^4 + ^4 )^2 Differentiating x and y with respect to : dx d = + = ( + ) dy d = 4 ^3 ( ) - 4 ^3 (- ) = 4 ( ^4 + ^4 ) Now, finding dy dx : dy dx = dy d dx d = 4 ( ^4 + ^4 ) ( + ) = 4( ^4 + ^4 ) + Squaring both sides: ( dy dx )^2 = 16( ^4 + ^4 )^2 ( + )^2 = 16(y^2+4) x^2+4 Rearranging the equation: (x^2+4) ( dy dx )^2 = 16(y^2+4) Differentiating both sides with respect to x: 2x ( dy dx )^2 + (x^2+4) 2 dy dx d^2y dx^2 = 16 2y dy dx Dividing the entire equation by 2 dy dx : x dy dx + (x^2+4) d^2y dx^2 = 16y (x^2+4) d^2y dx^2 - 16y = -x dy dx Substituting this into the given expression: ( x^2+4 y^2+4 ) dy dx [(x^2+4) d^2y dx^2 - 16y ] = ( x^2+4 y^2+4 ) dy dx (-x dy dx ) This simplifies to: -x ( x^2+4 y^2+4 ) ( dy dx )^2 Substituting ( dy dx )^2 = 16(y^2+4) x^2+4 into the expression: -x ( x^2+4 y^2+4 ) ( 16(y^2+4) x^2+4 ) = -16x Answer: -16x

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