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Calculus - Second Order Derivatives

Grade 12ICSE

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

πŸ”‘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 fβ€²β€²(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}, yβ€²β€²y'', y2y_2, fβ€²β€²(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 fβ€²β€²(x)>0f''(x) > 0 on an interval, the graph is 'concave up', resembling a cup or UU-shape. If fβ€²β€²(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, fβ€²β€²(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 fβ€²(c)=0f'(c) = 0 and fβ€²β€²(c)<0f''(c) < 0, the point x=cx=c is a local maximum (top of a hill). If fβ€²(c)=0f'(c) = 0 and fβ€²β€²(c)>0f''(c) > 0, the point x=cx=c is a local minimum (bottom of a valley). If fβ€²β€²(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 sβ€²(t)=vs'(t) = v represents velocity. The second derivative sβ€²β€²(t)=as''(t) = a represents acceleration, which is the rate of change of velocity over time.

πŸ“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)=uβ€²β€²v+2uβ€²vβ€²+uvβ€²β€²\frac{d^2}{dx^2}(uv) = u''v + 2u'v' + uv''

Second Derivative Test: fβ€²(c)=0Β andΒ fβ€²β€²(c)>0β€…β€ŠβŸΉβ€…β€ŠLocalΒ Minf'(c) = 0 \text{ and } f''(c) > 0 \implies \text{Local Min}

Second Derivative Test: fβ€²(c)=0Β andΒ fβ€²β€²(c)<0β€…β€ŠβŸΉβ€…β€ŠLocalΒ Maxf'(c) = 0 \text{ and } f''(c) < 0 \implies \text{Local Max}

πŸ’‘Examples

Problem 1:

If y=eaxsin⁑bxy = e^{ax} \sin bx, prove that d2ydx2βˆ’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(βˆ’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(βˆ’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[(a2βˆ’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[(a2βˆ’b2)sin⁑bx+2abcos⁑bx]βˆ’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⁑bxβˆ’b2sin⁑bx+2abcos⁑bxβˆ’2abcos⁑bxβˆ’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[(a2βˆ’b2βˆ’2a2+a2+b2)sin⁑bx+(2abβˆ’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(ΞΈ+sin⁑θ)x = a(\theta + \sin \theta) and y=a(1βˆ’cos⁑θ)y = a(1 - \cos \theta) at ΞΈ=Ο€2\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 ddΞΈ(dydx)β‹…dΞΈdx\frac{d}{d\theta}(\frac{dy}{dx}) \cdot \frac{d\theta}{dx}. ddΞΈ(tan⁑(ΞΈ/2))=12sec⁑2(ΞΈ/2)\frac{d}{d\theta}(\tan(\theta/2)) = \frac{1}{2} \sec^2(\theta/2) d2ydx2=(12sec⁑2(ΞΈ/2))β‹…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(ΞΈ/2)1 + \cos \theta = 2 \cos^2(\theta/2): d2ydx2=sec⁑2(ΞΈ/2)2β‹…12acos⁑2(ΞΈ/2)=14asec⁑4(ΞΈ/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 ΞΈ=Ο€2\theta = \frac{\pi}{2}. d2ydx2Β atΒ ΞΈ=Ο€2=14asec⁑4(Ο€/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.