More practice questions

5. Q5. Find d²y/dx² for (i) y = log (log x), x > 1; (ii) y = sin (log x), x > 0 (NCERT Exercise 5.7, Q9 and Q10, page 139).

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   (i) y₁ = 1/(x log x) = (x log x)⁻¹, so y₂ = −(1 + log x)/(x log x)². (ii) y₁ = cos (log x)/x, so y₂ = [−sin (log x) − cos (log x)]/x².

6. Q6. If y = (tan^⁻¹ x)², show that (x² + 1)²y₂ + 2x(x² + 1)y₁ = 2 (NCERT Exercise 5.7, Q17, page 140).

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   y₁ = 2 tan^⁻¹ x/(1 + x²), so (1 + x²)y₁ = 2 tan^⁻¹ x. Differentiating: (1 + x²)y₂ + 2xy₁ = 2/(1 + x²). Multiplying by 1 + x² gives the result.

7. Q7. Find d²y/dx² for the parabola x = at², y = 2at.

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   For a ≠ 0 and t ≠ 0, dy/dx = 1/t. Then d²y/dx² = [d(1/t)/dt] ÷ (dx/dt) = (−1/t²)/(2at) = −1/(2at³). Note that it is not (d²y/dt²)/(d²x/dt²), which would be 0.

8. Q8. For y = x³ − 6x² + 9x, find the values of x where dy/dx = 0, and the sign of d²y/dx² at each.

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   y₁ = 3x² − 12x + 9 = 3(x − 1)(x − 3), which is 0 at x = 1 and 3. y₂ = 6x − 12 is −6 at x = 1 and 6 at x = 3. (These signs are used to find maxima and minima in the next chapter.)