More practice questions

5. Q5. Differentiate sin (ax + b)/cos (cx + d) (NCERT Exercise 5.2, Q5, page 122).

   Show answerHide answer

   Write it as sin (ax + b) sec (cx + d) and use the product rule. The derivative is a cos (ax + b) sec (cx + d), plus c sin (ax + b) times tan (cx + d) sec (cx + d).

6. Q6. Find dy/dx if xy + y² = tan x + y (NCERT Exercise 5.3, Q4, page 125).

   Show answerHide answer

   y + x dy/dx + 2y dy/dx = sec² x + dy/dx, so dy/dx = (sec² x − y)/(x + 2y − 1).

7. Q7. Find dy/dx if x³ + x²y + xy² + y³ = 81 (NCERT Exercise 5.3, Q6, page 125).

   Show answerHide answer

   3x² + (2xy + x² dy/dx) + (y² + 2xy dy/dx) + 3y² dy/dx = 0, so dy/dx = −(3x² + 2xy + y²)/(x² + 2xy + 3y²).

8. Q8. Find dy/dx if sin² y + cos xy = κ, a constant (NCERT Exercise 5.3, Q7, page 125).

   Show answerHide answer

   2 sin y cos y dy/dx − sin xy (y + x dy/dx) = 0, so dy/dx = y sin xy/(sin 2y − x sin xy).

9. Q9. Find dy/dx for y = sin^⁻¹ ((1 − x²)/(1 + x²)), 0 < x < 1, and for y = cos^⁻¹ (2x/(1 + x²)), −1 < x < 1 (NCERT Exercise 5.3, Q12 and Q13, page 125).

   Show answerHide answer

   For the first, sin^⁻¹ u = π/2 − cos^⁻¹ u, and cos^⁻¹ ((1 − x²)/(1 + x²)) = 2 tan^⁻¹ x here, so y = π/2 − 2 tan^⁻¹ x and dy/dx = −2/(1 + x²). For the second, cos^⁻¹ u = π/2 − sin^⁻¹ u and sin^⁻¹ (2x/(1 + x²)) = 2 tan^⁻¹ x for −1 < x < 1, so dy/dx = −2/(1 + x²).

10. Q10. Find dy/dx for y = sin^⁻¹ (2x√(1 − x²)), −1/√2 < x < 1/√2 (NCERT Exercise 5.3, Q14, page 125).

    Show answerHide answer

    Put x = sin θ with −π/4 < θ < π/4. Then 2x√(1 − x²) = 2 sin θ cos θ = sin 2θ, with 2θ between −π/2 and π/2, so y = 2θ = 2 sin^⁻¹ x and dy/dx = 2/√(1 − x²).

11. Q11. Find dy/dx for y = sec^⁻¹ (1/(2x² − 1)), 0 < x < 1/√2 (NCERT Exercise 5.3, Q15, page 125).

    Show answerHide answer

    sec^⁻¹ (1/u) = cos^⁻¹ u, so y = cos^⁻¹ (2x² − 1). With x = cos θ, π/4 < θ < π/2, 2x² − 1 = cos 2θ with π/2 < 2θ < π, so y = 2θ = 2 cos^⁻¹ x and dy/dx = −2/√(1 − x²).

12. Q12. Is f(x) = x|x| differentiable at 0?

    Show answerHide answer

    (f(h) − f(0))/h = h|h|/h = |h|, which tends to 0 from both sides. So f is differentiable at 0, with f′(0) = 0, even though |x| itself is not.

13. Q13. Find a and b so that f(x) = x² for x ≤ 1 and ax + b for x > 1 is differentiable at 1.

    Show answerHide answer

    Continuity at 1 needs a + b = 1 = f(1). Then the right-hand derivative is the limit of (a(1 + h) + b − 1)/h, which is a, and the left-hand derivative is the slope of x² at 1, which is 2. So a = 2 and b = −1. Matching the slopes alone is not enough: with a = 2 and b = 0 the slopes of the two separate formulas match, but f jumps at 1 and is not even continuous there.