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Class 12 Continuity And Differentiability · Long Answer3 Markschain_and_product_rules

Class 12 Maths · Continuity And Differentiability · Long Answer #128

Differentiate y=\(\sqrt{x^{2} + 1} - \log{\left(\sqrt{1 + \frac{1}{x^{2}}} + \frac{1}{x} \right)}\) with respect to x. Assume x≠0.

✅ Step-by-Step Solution

Final Answer: \(\frac{x}{\sqrt{x^{2} + 1}} - \frac{- \frac{1}{x^{2}} - \frac{1}{x^{3} \sqrt{1 + \frac{1}{x^{2}}}}}{\sqrt{1 + \frac{1}{x^{2}}} + \frac{1}{x}}\)
  1. Use d√(x²+1)/dx=x/√(x²+1). For the logarithm put u=1/x, so ln(u+√(1+u²)) has derivative u′/√(1+u²)=−1/[x²√(1+1/x²)]. Subtract this derivative.
  2. Applying the indicated rules gives y′=\(\frac{x}{\sqrt{x^{2} + 1}} - \frac{- \frac{1}{x^{2}} - \frac{1}{x^{3} \sqrt{1 + \frac{1}{x^{2}}}}}{\sqrt{1 + \frac{1}{x^{2}}} + \frac{1}{x}}\).
  3. Collecting factors gives y′=\(\frac{x^{3} \sqrt{x^{2} + 1} + x^{2} \left|{x}\right| + x + \sqrt{x^{2} + 1} \left|{x}\right|}{x \sqrt{x^{2} + 1} \left(x \sqrt{x^{2} + 1} + \left|{x}\right|\right)}\).
  4. Therefore dy/dx=\(\frac{x}{\sqrt{x^{2} + 1}} - \frac{- \frac{1}{x^{2}} - \frac{1}{x^{3} \sqrt{1 + \frac{1}{x^{2}}}}}{\sqrt{1 + \frac{1}{x^{2}}} + \frac{1}{x}}\).

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Use d√(x²+1)/dx=x/√(x²+1). For the logarithm put u=1/x, so ln(u+√(1+u²)) has derivative u′/√(1+u²)=−1/[x²√(1+1/x²)]. Subtract this derivative.
  • [1 mark] Applying the indicated rules gives y′=\(\frac{x}{\sqrt{x^{2} + 1}} - \frac{- \frac{1}{x^{2}} - \frac{1}{x^{3} \sqrt{1 + \frac{1}{x^{2}}}}}{\sqrt{1 + \frac{1}{x^{2}}} + \frac{1}{x}}\).
  • [1 mark] Collecting factors gives y′=\(\frac{x^{3} \sqrt{x^{2} + 1} + x^{2} \left|{x}\right| + x + \sqrt{x^{2} + 1} \left|{x}\right|}{x \sqrt{x^{2} + 1} \left(x \sqrt{x^{2} + 1} + \left|{x}\right|\right)}\).
  • [0.5 marks] Therefore dy/dx=\(\frac{x}{\sqrt{x^{2} + 1}} - \frac{- \frac{1}{x^{2}} - \frac{1}{x^{3} \sqrt{1 + \frac{1}{x^{2}}}}}{\sqrt{1 + \frac{1}{x^{2}}} + \frac{1}{x}}\).

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