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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

If _ 10 ( x^ 3 - y^ 3 x^ 3 + y^ 3 ) = 2 , then (dy)/(dx) =

If log⁡10(x3−y3x3+y3)=2\log_{10} \left(\dfrac{x^{3} - y^{3}}{x^{3} + y^{3}}\right) = 2, then dydx=\dfrac{dy}{dx} =

Options

  1. Axy\dfrac{x}{y}
  2. B−yx- \dfrac{y}{x}
  3. C−xy- \dfrac{x}{y}
  4. Dyx\dfrac{y}{x}

Correct answer

D. yx\dfrac{y}{x}

Step-by-step solution

Given,& log⁡10(x3−y3x3+y3)=2\log_{10} \left(\dfrac{x^{3} - y^{3}}{x^{3} + y^{3}}\right) = 2

⇒(x3−y3x3+y3)=(10)2=100⇒ \left(\dfrac{x^{3} - y^{3}}{x^{3} + y^{3}}\right) = \left(10\right)2 = 100

Applying componendo and dividendo, we get,

x3y3=−10199⇒xy=(−10199)13\dfrac{x3}{y3} = - \dfrac{101}{99} ⇒ \dfrac{x}{y} = \left(- \dfrac{101}{99}\right)\dfrac{1}{3}......(i)

Hence, y=(−99101)13xy = \left(- \dfrac{99}{101}\right)\dfrac{1}{3}x

Differentiating, we get,

dydx=(−99101)13=yx\dfrac{dy}{dx} = \left(- \dfrac{99}{101}\right)\dfrac{1}{3} = \dfrac{y}{x}& (from (i))
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