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Class 12 Continuity And Differentiability · Long Answer3 Marksparametric_derivatives

Class 12 Maths · Continuity And Differentiability · Long Answer #81

For x=\(e^{\cos{\left(2 t \right)}}\), y=\(e^{\sin{\left(2 t \right)}}\), find dy/dx. Assume sin2t≠0. Express the result using x,y and their natural logarithms.

✅ Step-by-Step Solution

Final Answer: −y ln x/(x ln y)
  1. Differentiate in t: dx/dt=\(- 2 e^{\cos{\left(2 t \right)}} \sin{\left(2 t \right)}\) and dy/dt=\(2 e^{\sin{\left(2 t \right)}} \cos{\left(2 t \right)}\).
  2. Thus dy/dx=(dy/dt)/(dx/dt)=\(- \frac{e^{\sin{\left(2 t \right)}} e^{- \cos{\left(2 t \right)}}}{\tan{\left(2 t \right)}}\).
  3. Simplify the ratio on the stated regular parameter intervals to \(- \frac{e^{\sin{\left(2 t \right)}} e^{- \cos{\left(2 t \right)}}}{\tan{\left(2 t \right)}}\).
  4. Since ln x=cos2t and ln y=sin2t, the ratio is dy/dx=−y ln x/(x ln y).

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Differentiate in t: dx/dt=\(- 2 e^{\cos{\left(2 t \right)}} \sin{\left(2 t \right)}\) and dy/dt=\(2 e^{\sin{\left(2 t \right)}} \cos{\left(2 t \right)}\).
  • [1 mark] Thus dy/dx=(dy/dt)/(dx/dt)=\(- \frac{e^{\sin{\left(2 t \right)}} e^{- \cos{\left(2 t \right)}}}{\tan{\left(2 t \right)}}\).
  • [1 mark] Simplify the ratio on the stated regular parameter intervals to \(- \frac{e^{\sin{\left(2 t \right)}} e^{- \cos{\left(2 t \right)}}}{\tan{\left(2 t \right)}}\).
  • [0.5 marks] Since ln x=cos2t and ln y=sin2t, the ratio is dy/dx=−y ln x/(x ln y).

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