Class 12 Continuity And Differentiability · Short Answer2 Marksparametric_derivatives

Class 12 Maths · Continuity And Differentiability · Short Answer #53

For x=\(a \sin{\left(2 t \right)}\), y=\(a \left(\log{\left(\tan{\left(t \right)} \right)} + \cos{\left(2 t \right)}\right)\), find dy/dx. Assume a≠0, tan t>0, sin2t≠0 and cos2t≠0.

✅ Step-by-Step Solution

Final Answer: \(\frac{1}{\tan{\left(2 t \right)}}\)

Differentiate in t: dx/dt=\(2 a \cos{\left(2 t \right)}\) and dy/dt=\(a \left(\frac{\tan^{2}{\left(t \right)} + 1}{\tan{\left(t \right)}} - 2 \sin{\left(2 t \right)}\right)\).
Thus dy/dx=(dy/dt)/(dx/dt)=\(\frac{1}{\tan{\left(2 t \right)}}\).
Therefore dy/dx=\(\frac{1}{\tan{\left(2 t \right)}}\).

📋 CBSE Stepwise Marking Scheme

[0.5 marks] Differentiate in t: dx/dt=\(2 a \cos{\left(2 t \right)}\) and dy/dt=\(a \left(\frac{\tan^{2}{\left(t \right)} + 1}{\tan{\left(t \right)}} - 2 \sin{\left(2 t \right)}\right)\).
[1 mark] Thus dy/dx=(dy/dt)/(dx/dt)=\(\frac{1}{\tan{\left(2 t \right)}}\).
[0.5 marks] Therefore dy/dx=\(\frac{1}{\tan{\left(2 t \right)}}\).

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Class 12 Maths · Continuity And Differentiability · Short Answer #53

✅ Step-by-Step Solution

📋 CBSE Stepwise Marking Scheme

Master Chapter 5: Continuity And Differentiability

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