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Calculus - Second Order Derivatives

Grade 12ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition of Successive Differentiation: If a function y=f(x)y = f(x) is differentiable with respect to xx, its derivative dydx\frac{dy}{dx} is called the first-order derivative. If dydx\frac{dy}{dx} is also differentiable, its derivative with respect to xx is called the second-order derivative of yy with respect to xx, denoted as d2ydx2\frac{d^2y}{dx^2} or f′′(x)f''(x).

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Notational Variations: The second-order derivative can be represented in multiple ways depending on the context: d2ydx2\frac{d^2y}{dx^2}, y′′y'', y2y_2, f′′(x)f''(x), or D2yD^2y. In ICSE problems, y2y_2 is frequently used in differential equations to represent the second derivative.

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Geometric Interpretation and Concavity: Visually, the second derivative represents the 'curvature' of a graph. If f′′(x)>0f''(x) > 0 on an interval, the graph is 'concave up', resembling a cup or UU-shape. If f′′(x)<0f''(x) < 0, the graph is 'concave down', resembling a cap or an inverted UU-shape.

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Points of Inflection: A point on the curve where the concavity changes (from concave up to concave down, or vice versa) is called a point of inflection. At such a point, f′′(x)=0f''(x) = 0 or is undefined, provided the tangent exists. Visually, this is where the curve transitions from curving upwards to curving downwards.

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Second Derivative Test for Local Extrema: This test helps identify if a critical point is a maximum or minimum. If f′(c)=0f'(c) = 0 and f′′(c)<0f''(c) < 0, the point x=cx=c is a local maximum (top of a hill). If f′(c)=0f'(c) = 0 and f′′(c)>0f''(c) > 0, the point x=cx=c is a local minimum (bottom of a valley). If f′′(c)=0f''(c) = 0, the test is inconclusive.

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Parametric Second Derivatives: For functions defined parametrically as x=f(t)x = f(t) and y=g(t)y = g(t), finding the second derivative requires care. First find dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, then differentiate this result with respect to tt and multiply by dtdx\frac{dt}{dx} using the chain rule. Visually, this describes the change in slope along a path defined by a third variable.

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Physical Significance in Kinematics: If s(t)s(t) represents the displacement of a particle at time tt, then the first derivative s′(t)=vs'(t) = v represents velocity. The second derivative s′′(t)=as''(t) = a represents acceleration, which is the rate of change of velocity over time.

📐Formulae

d2ydx2=ddx(dydx)\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)

y2=ddx(y1)y_2 = \frac{d}{dx}(y_1)

Parametric Formula: d2ydx2=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}

Product Rule for Second Derivative: d2dx2(uv)=u′′v+2u′v′+uv′′\frac{d^2}{dx^2}(uv) = u''v + 2u'v' + uv''

Second Derivative Test: f′(c)=0 and f′′(c)>0  ⟹  Local Minf'(c) = 0 \text{ and } f''(c) > 0 \implies \text{Local Min}

Second Derivative Test: f′(c)=0 and f′′(c)<0  ⟹  Local Maxf'(c) = 0 \text{ and } f''(c) < 0 \implies \text{Local Max}

💡Examples

Problem 1:

If y=eaxsin⁡bxy = e^{ax} \sin bx, prove that d2ydx2−2adydx+(a2+b2)y=0\frac{d^2y}{dx^2} - 2a \frac{dy}{dx} + (a^2 + b^2)y = 0.

Solution:

Step 1: Find the first derivative dydx\frac{dy}{dx} using the product rule. dydx=eax(bcos⁡bx)+aeaxsin⁡bx=eax(bcos⁡bx+asin⁡bx)\frac{dy}{dx} = e^{ax}(b \cos bx) + a e^{ax} \sin bx = e^{ax}(b \cos bx + a \sin bx)

Step 2: Find the second derivative d2ydx2\frac{d^2y}{dx^2} by differentiating again. d2ydx2=eax(−b2sin⁡bx+abcos⁡bx)+aeax(bcos⁡bx+asin⁡bx)\frac{d^2y}{dx^2} = e^{ax}(-b^2 \sin bx + ab \cos bx) + a e^{ax}(b \cos bx + a \sin bx) d2ydx2=eax(−b2sin⁡bx+abcos⁡bx+abcos⁡bx+a2sin⁡bx)\frac{d^2y}{dx^2} = e^{ax}(-b^2 \sin bx + ab \cos bx + ab \cos bx + a^2 \sin bx) d2ydx2=eax[(a2−b2)sin⁡bx+2abcos⁡bx]\frac{d^2y}{dx^2} = e^{ax}[(a^2 - b^2) \sin bx + 2ab \cos bx]

Step 3: Substitute d2ydx2\frac{d^2y}{dx^2}, dydx\frac{dy}{dx}, and yy into the LHS of the equation. LHS=eax[(a2−b2)sin⁡bx+2abcos⁡bx]−2a[eax(bcos⁡bx+asin⁡bx)]+(a2+b2)eaxsin⁡bxLHS = e^{ax}[(a^2 - b^2) \sin bx + 2ab \cos bx] - 2a[e^{ax}(b \cos bx + a \sin bx)] + (a^2 + b^2)e^{ax} \sin bx Taking eaxe^{ax} common: =eax[a2sin⁡bx−b2sin⁡bx+2abcos⁡bx−2abcos⁡bx−2a2sin⁡bx+a2sin⁡bx+b2sin⁡bx]= e^{ax} [a^2 \sin bx - b^2 \sin bx + 2ab \cos bx - 2ab \cos bx - 2a^2 \sin bx + a^2 \sin bx + b^2 \sin bx] =eax[(a2−b2−2a2+a2+b2)sin⁡bx+(2ab−2ab)cos⁡bx]= e^{ax} [ (a^2 - b^2 - 2a^2 + a^2 + b^2) \sin bx + (2ab - 2ab) \cos bx ] =eax[0+0]=0=RHS= e^{ax} [0 + 0] = 0 = RHS.

Explanation:

The solution involves successive differentiation and substitution into the given differential equation. The product rule is applied twice to manage the exponential and trigonometric components.

Problem 2:

Find d2ydx2\frac{d^2y}{dx^2} for the parametric equations x=a(θ+sin⁡θ)x = a(\theta + \sin \theta) and y=a(1−cos⁡θ)y = a(1 - \cos \theta) at θ=π2\theta = \frac{\pi}{2}.

Solution:

Step 1: Differentiate xx and yy with respect to θ\theta. dxdθ=a(1+cos⁡θ)\frac{dx}{d\theta} = a(1 + \cos \theta) dydθ=a(sin⁡θ)\frac{dy}{d\theta} = a(\sin \theta)

Step 2: Find dydx\frac{dy}{dx}. dydx=dy/dθdx/dθ=asin⁡θa(1+cos⁡θ)=2sin⁡(θ/2)cos⁡(θ/2)2cos⁡2(θ/2)=tan⁡(θ/2)\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \sin \theta}{a(1 + \cos \theta)} = \frac{2 \sin(\theta/2) \cos(\theta/2)}{2 \cos^2(\theta/2)} = \tan(\theta/2)

Step 3: Find d2ydx2\frac{d^2y}{dx^2} using the formula ddθ(dydx)⋅dθdx\frac{d}{d\theta}(\frac{dy}{dx}) \cdot \frac{d\theta}{dx}. ddθ(tan⁡(θ/2))=12sec⁡2(θ/2)\frac{d}{d\theta}(\tan(\theta/2)) = \frac{1}{2} \sec^2(\theta/2) d2ydx2=(12sec⁡2(θ/2))⋅1a(1+cos⁡θ)\frac{d^2y}{dx^2} = \left(\frac{1}{2} \sec^2(\theta/2)\right) \cdot \frac{1}{a(1 + \cos \theta)} Since 1+cos⁡θ=2cos⁡2(θ/2)1 + \cos \theta = 2 \cos^2(\theta/2): d2ydx2=sec⁡2(θ/2)2⋅12acos⁡2(θ/2)=14asec⁡4(θ/2)\frac{d^2y}{dx^2} = \frac{\sec^2(\theta/2)}{2} \cdot \frac{1}{2a \cos^2(\theta/2)} = \frac{1}{4a} \sec^4(\theta/2)

Step 4: Evaluate at θ=π2\theta = \frac{\pi}{2}. d2ydx2 at θ=π2=14asec⁡4(π/4)=14a(2)4=44a=1a\frac{d^2y}{dx^2} \text{ at } \theta = \frac{\pi}{2} = \frac{1}{4a} \sec^4(\pi/4) = \frac{1}{4a} (\sqrt{2})^4 = \frac{4}{4a} = \frac{1}{a}.

Explanation:

This problem demonstrates the specific chain rule required for second derivatives of parametric functions. It also utilizes trigonometric identities to simplify the first derivative before the second differentiation.