Class 12 Maths: Continuity and Differentiability Ex 5.7 - Second Order Derivatives

Understanding Second Order Derivatives

You already know that for a function y = f(x), its first derivative, denoted as dy/dx or f'(x), represents the instantaneous rate of change of y with respect to x. Geometrically, it gives us the slope of the tangent to the curve at any point. Now, imagine that this first derivative dy/dx is itself a function of x. If dy/dx is differentiable, we can differentiate it again with respect to x. This new derivative, derived from the first derivative, is what we call the second-order derivative.

The notation for the second-order derivative is d/dx (dy/dx), which is compactly written as d²y/dx² (read as "d two y by dx squared"). Other common notations include f''(x) (f double prime of x) or simply y''. These notations all mean the same thing: differentiate the function twice with respect to x.

Why is this important? The second derivative provides crucial insights:

• Physical Interpretation: If s = f(t) is the displacement of an object at time t, then ds/dt is its velocity v(t). Consequently, d²s/dt² represents the acceleration a(t) of the object. It tells us how the velocity is changing.
• Geometric Interpretation: While the first derivative tells us if a function is increasing or decreasing, the second derivative tells us about the concavity of the curve. If f''(x) > 0, the curve is concave up (it opens upwards, like a cup). If f''(x) < 0, the curve is concave down (it opens downwards, like an inverted cup). This is fundamental for sketching graphs and finding local extrema.

Step-by-Step Procedure to Find Second Order Derivatives

1. Step 1: Find the First Derivative (dy/dx) — Start by differentiating the given function y = f(x) with respect to x. Apply all standard differentiation rules (power rule, product rule, quotient rule, chain rule, derivatives of trigonometric, exponential, and logarithmic functions) as needed. This will give you dy/dx or f'(x).
2. Step 2: Differentiate the First Derivative Again (d²y/dx²) — Treat the expression you obtained in Step 1 (which is dy/dx) as a new function of x. Now, differentiate this new function with respect to x once more. Again, apply all necessary differentiation rules carefully. The result will be the second-order derivative, d²y/dx² or f''(x).

Worked Examples: Calculating Second Order Derivatives

• Example 1: Find d²y/dx² for y = x⁴ + 3x³ - 5x + 10. Step 1: Find the first derivative (dy/dx) Given y = x⁴ + 3x³ - 5x + 10 dy/dx = d/dx (x⁴) + d/dx (3x³) - d/dx (5x) + d/dx (10) dy/dx = 4x³ + 3(3x²) - 5(1) + 0 dy/dx = 4x³ + 9x² - 5 Step 2: Differentiate dy/dx again (d²y/dx²) Now, differentiate 4x³ + 9x² - 5 with respect to x: d²y/dx² = d/dx (4x³ + 9x² - 5) d²y/dx² = d/dx (4x³) + d/dx (9x²) - d/dx (5) d²y/dx² = 4(3x²) + 9(2x) - 0 d²y/dx² = 12x² + 18x Final Answer: d²y/dx² = 12x² + 18x
• Example 2: Find d²y/dx² for y = e^(2x) sin(3x). Step 1: Find the first derivative (dy/dx) Given y = e^(2x) sin(3x). We use the product rule: d/dx (uv) = u'v + uv' Let u = e^(2x) and v = sin(3x). Then u' = d/dx (e^(2x)) = e^(2x) d/dx(2x) = 2e^(2x) (using chain rule). And v' = d/dx (sin(3x)) = cos(3x) d/dx(3x) = 3cos(3x) (using chain rule). dy/dx = u'v + uv' = (2e^(2x))sin(3x) + e^(2x)(3cos(3x)) dy/dx = e^(2x) [2sin(3x) + 3cos(3x)] Step 2: Differentiate dy/dx again (d²y/dx²) Now, differentiate e^(2x) [2sin(3x) + 3cos(3x)] using the product rule again. Let U = e^(2x) and V = 2sin(3x) + 3cos(3x). U' = 2e^(2x) (from previous step). V' = d/dx (2sin(3x) + 3cos(3x)) V' = 2(cos(3x)3) + 3(-sin(3x)3) V' = 6cos(3x) - 9sin(3x) d²y/dx² = U'V + UV' d²y/dx² = (2e^(2x))[2sin(3x) + 3cos(3x)] + e^(2x)[6cos(3x) - 9sin(3x)] Factor out e^(2x): d²y/dx² = e^(2x) [2(2sin(3x) + 3cos(3x)) + (6cos(3x) - 9sin(3x))] d²y/dx² = e^(2x) [4sin(3x) + 6cos(3x) + 6cos(3x) - 9sin(3x)] d²y/dx² = e^(2x) [-5sin(3x) + 12cos(3x)] Final Answer: d²y/dx² = e^(2x) [12cos(3x) - 5sin(3x)]
• Example 3: If y = A e^(mx) + B e^(nx), show that d²y/dx² - (m+n)dy/dx + mny = 0. Step 1: Find dy/dx Given y = A e^(mx) + B e^(nx) dy/dx = d/dx (A e^(mx)) + d/dx (B e^(nx)) dy/dx = A (e^(mx) m) + B (e^(nx) n) dy/dx = Am e^(mx) + Bn e^(nx) Step 2: Find d²y/dx² d²y/dx² = d/dx (Am e^(mx) + Bn e^(nx)) d²y/dx² = Am (e^(mx) m) + Bn (e^(nx) n) d²y/dx² = Am² e^(mx) + Bn² e^(nx) Step 3: Substitute into the given equation We need to show d²y/dx² - (m+n)dy/dx + mny = 0. LHS = (Am² e^(mx) + Bn² e^(nx)) - (m+n)(Am e^(mx) + Bn e^(nx)) + mn(A e^(mx) + B e^(nx)) Expand the terms: LHS = Am² e^(mx) + Bn² e^(nx) - (Am² e^(mx) + Amn e^(mx) + Bmn e^(nx) + Bn² e^(nx)) + (Amn e^(mx) + Bmn e^(nx)) Combine like terms (terms with e^(mx) and e^(nx)): For e^(mx) terms: Am² - Am² - Amn + Amn = 0 For e^(nx) terms: Bn² - Bmn - Bn² + Bmn = 0 Since both sets of terms sum to zero, the LHS = 0. Final Answer: LHS = 0 = RHS. Hence proved.

Exam Tips and Common Mistakes to Avoid

To ace questions on second-order derivatives, be mindful of these points:

• Chain Rule is Crucial: Many errors occur when students forget to apply the chain rule correctly during the second differentiation, especially with composite functions like sin(2x) or e^(3x). Remember, d/dx(f(g(x))) = f'(g(x)) * g'(x).
• Distinguish (dy/dx)² from d²y/dx²: These are NOT the same! (dy/dx)² means the square of the first derivative, while d²y/dx² is the second derivative. Do not interchange them.
• Algebraic Accuracy: Be very careful with signs and algebraic simplifications. A small mistake in simplification can lead to an entirely incorrect final answer, especially in proofs.
• Implicit Differentiation: If y is an implicit function of x, remember to differentiate y terms with respect to x using the chain rule (e.g., d/dx(y²) = 2y(dy/dx)). The same applies when finding the second derivative: d/dx(dy/dx) might involve differentiating terms that still contain dy/dx.
• Practice Proofs: Questions asking you to prove a relation involving y, dy/dx, and d²y/dx² are common. Calculate each derivative separately, then substitute them into the given equation and simplify the Left Hand Side (LHS) to show it equals the Right Hand Side (RHS).

Practice Questions with Solutions

• Q: Find d²y/dx² for y = x² cos x. A: Step 1: Find dy/dx using the product rule. Let u = x², v = cos x. Then u' = 2x, v' = -sin x. dy/dx = u'v + uv' = (2x)(cos x) + (x²)(-sin x) = 2x cos x - x² sin x. Step 2: Find d²y/dx² by differentiating dy/dx again. This requires the product rule for both terms. For 2x cos x: d/dx(2x cos x) = 2 cos x + 2x(-sin x) = 2 cos x - 2x sin x. For -x² sin x: d/dx(-x² sin x) = -(2x sin x + x² cos x) = -2x sin x - x² cos x. Add the results: d²y/dx² = (2 cos x - 2x sin x) + (-2x sin x - x² cos x) d²y/dx² = 2 cos x - 4x sin x - x² cos x. Final answer: d²y/dx² = (2 - x²)cos x - 4x sin x.
• Q: If y = 3e^(2x) + 2e^(3x), show that d²y/dx² - 5dy/dx + 6y = 0. A: Step 1: Find dy/dx. dy/dx = d/dx(3e^(2x) + 2e^(3x)) = 3(e^(2x)2) + 2(e^(3x)3) = 6e^(2x) + 6e^(3x). Step 2: Find d²y/dx². d²y/dx² = d/dx(6e^(2x) + 6e^(3x)) = 6(e^(2x)2) + 6(e^(3x)3) = 12e^(2x) + 18e^(3x). Step 3: Substitute y, dy/dx, and d²y/dx² into the given equation. LHS = (12e^(2x) + 18e^(3x)) - 5(6e^(2x) + 6e^(3x)) + 6(3e^(2x) + 2e^(3x)) LHS = 12e^(2x) + 18e^(3x) - 30e^(2x) - 30e^(3x) + 18e^(2x) + 12e^(3x) Group terms with e^(2x): (12 - 30 + 18)e^(2x) = 0e^(2x) = 0. Group terms with e^(3x): (18 - 30 + 12)e^(3x) = 0e^(3x) = 0. LHS = 0 + 0 = 0. Final answer: LHS = 0 = RHS. Hence proved.
• Q: Find d²y/dx² if y = sin(log x). A: Step 1: Find dy/dx using the chain rule. Let u = log x, so y = sin u. dy/du = cos u, du/dx = 1/x. dy/dx = (dy/du) (du/dx) = cos(log x) (1/x) = (cos(log x))/x. Step 2: Find d²y/dx² using the quotient rule for (cos(log x))/x. Let u = cos(log x) and v = x. du/dx = -sin(log x) (1/x) (using chain rule) dv/dx = 1. d²y/dx² = (v (du/dx) - u (dv/dx)) / v² d²y/dx² = (x (-sin(log x)/x) - cos(log x) * 1) / x² d²y/dx² = (-sin(log x) - cos(log x)) / x². Final answer: d²y/dx² = -(sin(log x) + cos(log x)) / x².
• Q: If y = 5cos x - 3sin x, prove that d²y/dx² + y = 0. A: Step 1: Find dy/dx. dy/dx = d/dx(5cos x - 3sin x) = 5(-sin x) - 3(cos x) = -5sin x - 3cos x. Step 2: Find d²y/dx². d²y/dx² = d/dx(-5sin x - 3cos x) = -5(cos x) - 3(-sin x) = -5cos x + 3sin x. Step 3: Substitute y and d²y/dx² into the given equation. LHS = d²y/dx² + y = (-5cos x + 3sin x) + (5cos x - 3sin x) LHS = (-5cos x + 5cos x) + (3sin x - 3sin x) LHS = 0 + 0 = 0. Final answer: LHS = 0 = RHS. Hence proved.

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