Class 12 Continuity And Differentiability · Long Answer3 Marksimplicit_differentiation
Class 12 Maths · Continuity And Differentiability · Long Answer #68
If y^x=e^(y−x) with y>0 and ln y≠0, show y′=(1+ln y)²/ln y.
✅ Step-by-Step Solution
Final Answer: (1+ln y)²/ln y
- Taking natural logarithms gives x ln y=y−x, hence x(1+ln y)=y. On this relation 1+ln y cannot be zero because y>0.
- Differentiate: ln y+(x/y)y′=y′−1. Thus (1−x/y)y′=1+ln y.
- The relation gives x/y=1/(1+ln y), so 1−x/y=ln y/(1+ln y). Dividing, with ln y≠0, yields the stated quotient.
- Therefore y′=(1+ln y)²/ln y.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Taking natural logarithms gives x ln y=y−x, hence x(1+ln y)=y. On this relation 1+ln y cannot be zero because y>0.
- [1 mark] Differentiate: ln y+(x/y)y′=y′−1. Thus (1−x/y)y′=1+ln y.
- [1 mark] The relation gives x/y=1/(1+ln y), so 1−x/y=ln y/(1+ln y). Dividing, with ln y≠0, yields the stated quotient.
- [0.5 marks] Therefore y′=(1+ln y)²/ln y.
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