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Class 12 Continuity And Differentiability · Long Answer3 Marksimplicit_differentiation

Class 12 Maths · Continuity And Differentiability · Long Answer #150

If (ax+b)e^(y/x)=x on a differentiable real branch with x≠0 and x/(ax+b)>0, show x³y″=(xy′−y)².

✅ Step-by-Step Solution

Final Answer: x³y″=(xy′−y)²
  1. The relation gives y=x ln[x/(ax+b)]. Write d=ax+b, which is nonzero on the stated domain.
  2. Product differentiation gives y′=ln(x/d)+1−ax/d=ln(x/d)+b/d. Hence xy′−y=bx/d.
  3. Differentiate again: y″=1/x−a/d−ab/d². Combining over xd² gives y″=b²/(xd²).
  4. Thus x³y″=b²x²/d²=(bx/d)²=(xy′−y)², proving the claim.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] The relation gives y=x ln[x/(ax+b)]. Write d=ax+b, which is nonzero on the stated domain.
  • [1 mark] Product differentiation gives y′=ln(x/d)+1−ax/d=ln(x/d)+b/d. Hence xy′−y=bx/d.
  • [1 mark] Differentiate again: y″=1/x−a/d−ab/d². Combining over xd² gives y″=b²/(xd²).
  • [0.5 marks] Thus x³y″=b²x²/d²=(bx/d)²=(xy′−y)², proving the claim.

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