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Differentiation question

2020 · 7 Jan · Shift 1 · Q28
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  5. /2020 · 7 Jan · Shift 1 · Q28

Differentiation question

2020 · 7 Jan · Shift 1 · Q28

JEE MainMathematicsDifferentiationMCQ+4 / −1
Let xk + yk = ak, (a, k > 0 ) and dydx+(yx)13=0{{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0dxdy​+(xy​)31​=0, then k is:
  1. A
    13{1 \over 3}31​
  2. B
    23{2 \over 3}32​
  3. C
    43{4 \over 3}34​
  4. D
    32{3 \over 2}23​
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