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If x = sin 2θ and y = cos 2θ, find dy/dx and d²y/dx².

Mathematics

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Graph of y = cos(2θ)

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Key points of y = cos(2θ)
Pointθy
Maximum−3.141
x-intercept−2.360
Minimum−1.57−1
x-intercept−0.790
Maximum01
x-intercept0.790
Minimum1.57−1
x-intercept2.360
Maximum3.141

Step-by-step solution

6 steps

7. (ii)

  1. Find dxdθ\frac{dx}{d\theta} and dydθ\frac{dy}{d\theta}.

    Given x=sin⁡2θx = \sin 2\theta and y=cos⁡2θy = \cos 2\theta.

    dxdθ=ddθ(sin⁡2θ)=2cos⁡2θ\frac{dx}{d\theta} = \frac{d}{d\theta}(\sin 2\theta) = 2 \cos 2\theta dydθ=ddθ(cos⁡2θ)=−2sin⁡2θ\frac{dy}{d\theta} = \frac{d}{d\theta}(\cos 2\theta) = -2 \sin 2\theta

  2. Find dydx\frac{dy}{dx} using the chain rule.

    dydx=dy/dθdx/dθ=−2sin⁡2θ2cos⁡2θ\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-2 \sin 2\theta}{2 \cos 2\theta} dydx=−sin⁡2θcos⁡2θ=−tan⁡2θ\frac{dy}{dx} = -\frac{\sin 2\theta}{\cos 2\theta} = -\tan 2\theta

  3. Find d2ydx2\frac{d^2y}{dx^2} using the chain rule.

    d2ydx2=ddθ(dydx)⋅1dx/dθ\frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \cdot \frac{1}{dx/d\theta}

    First, find ddθ(dydx)\frac{d}{d\theta}\left(\frac{dy}{dx}\right):

    ddθ(−tan⁡2θ)=−sec⁡2(2θ)⋅ddθ(2θ)\frac{d}{d\theta}(-\tan 2\theta) = -\sec^2(2\theta) \cdot \frac{d}{d\theta}(2\theta) =−sec⁡2(2θ)⋅2= -\sec^2(2\theta) \cdot 2 =−2sec⁡2(2θ)= -2 \sec^2(2\theta)

    Now substitute this back into the formula for d2ydx2\frac{d^2y}{dx^2}:

    d2ydx2=(−2sec⁡2(2θ))⋅12cos⁡2θ\frac{d^2y}{dx^2} = (-2 \sec^2(2\theta)) \cdot \frac{1}{2 \cos 2\theta} d2ydx2=−2sec⁡2(2θ)2cos⁡2θ\frac{d^2y}{dx^2} = \frac{-2 \sec^2(2\theta)}{2 \cos 2\theta}

    Since sec⁡2θ=1cos⁡2θ\sec 2\theta = \frac{1}{\cos 2\theta}:

    d2ydx2=−1cos⁡2(2θ)cos⁡2θ\frac{d^2y}{dx^2} = \frac{-1}{\cos^2(2\theta) \cos 2\theta} d2ydx2=−1cos⁡3(2θ)\frac{d^2y}{dx^2} = \frac{-1}{\cos^3(2\theta)} d2ydx2=−sec⁡3(2θ)\frac{d^2y}{dx^2} = \boxed{-\sec^3(2\theta)}

    7. (iii)

  4. Find the first derivative of the function f(x)f(x).

    Given f(x)=(2x−5)3+xf(x) = (2x-5)^3 + x.

    f′(x)=ddx((2x−5)3)+ddx(x)f'(x) = \frac{d}{dx}((2x-5)^3) + \frac{d}{dx}(x)

    Using the chain rule for (2x−5)3(2x-5)^3:

    ddx((2x−5)3)=3(2x−5)2⋅ddx(2x−5)\frac{d}{dx}((2x-5)^3) = 3(2x-5)^2 \cdot \frac{d}{dx}(2x-5) =3(2x−5)2⋅2= 3(2x-5)^2 \cdot 2 =6(2x−5)2= 6(2x-5)^2

    So, the derivative f′(x)f'(x) is:

    f′(x)=6(2x−5)2+1f'(x) = 6(2x-5)^2 + 1

  5. Analyze the sign of f′(x)f'(x).

    For any real number xx, the term (2x−5)2(2x-5)^2 is always greater than or equal to zero because it is a square.

    (2x−5)2≥0(2x-5)^2 \ge 0

    Multiplying by 6, the term remains non-negative:

    6(2x−5)2≥06(2x-5)^2 \ge 0

    Adding 1 to this expression:

    6(2x−5)2+1≥16(2x-5)^2 + 1 \ge 1

    Therefore, f′(x)≥1f'(x) \ge 1 for all x∈Rx \in \mathbb{R}.

  6. Conclude that the function is increasing.

    Since f′(x)≥1f'(x) \ge 1, it means f′(x)>0f'(x) > 0 for all x∈Rx \in \mathbb{R}. A function is increasing if its first derivative is always positive.
    Thus, the function f(x)=(2x−5)3+xf(x) = (2x-5)^3 + x is an increasing function.

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