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Class 12 Continuity And Differentiability · Long Answer3 Markssecond_derivatives

Class 12 Maths · Continuity And Differentiability · Long Answer #95

For y=\(\log{\left(x + \sqrt{a^{2} + x^{2}} \right)}\), prove \(a^{2} + x^{2}\)y″ + \(x\)y′ + \(0\)y = \(0\). Assume a≠0 (so the logarithm argument is positive for every real x).

✅ Step-by-Step Solution

Final Answer: Identity proved.
  1. Differentiate y=\(\log{\left(x + \sqrt{a^{2} + x^{2}} \right)}\) to obtain y′=\(\frac{1}{\sqrt{a^{2} + x^{2}}}\).
  2. Differentiate again, using the product and chain rules: y″=\(\frac{a^{2} \left(a^{2} + x^{2}\right) - \left(a^{2} + x^{2}\right)^{\frac{3}{2}} \left(x + \sqrt{a^{2} + x^{2}}\right)}{\left(a^{2} + x^{2}\right)^{\frac{5}{2}} \left(x + \sqrt{a^{2} + x^{2}}\right)}\).
  3. Substitute these derivatives into the left side \(a^{2} + x^{2}\)y″ + \(x\)y′ + \(0\)y = \(0\). Collecting like terms gives \(0\).
  4. This equals \(0\), proving the identity on the stated domain.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Differentiate y=\(\log{\left(x + \sqrt{a^{2} + x^{2}} \right)}\) to obtain y′=\(\frac{1}{\sqrt{a^{2} + x^{2}}}\).
  • [1 mark] Differentiate again, using the product and chain rules: y″=\(\frac{a^{2} \left(a^{2} + x^{2}\right) - \left(a^{2} + x^{2}\right)^{\frac{3}{2}} \left(x + \sqrt{a^{2} + x^{2}}\right)}{\left(a^{2} + x^{2}\right)^{\frac{5}{2}} \left(x + \sqrt{a^{2} + x^{2}}\right)}\).
  • [1 mark] Substitute these derivatives into the left side \(a^{2} + x^{2}\)y″ + \(x\)y′ + \(0\)y = \(0\). Collecting like terms gives \(0\).
  • [0.5 marks] This equals \(0\), proving the identity on the stated domain.

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