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Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Derivatives of Sin, Cos and Tan

The derivative of sin⁡x\sin x is cos⁡x\cos x, and the derivative of cos⁡x\cos x is −sin⁡x-\sin x. Every other trig derivative in MCV4U comes from those two and the rules you already know.

How It Works

The three rules.

ddxsin⁡x=cos⁡x,ddxcos⁡x=−sin⁡x,ddxtan⁡x=sec⁡2x\frac{d}{dx}\sin x = \cos x, \qquad \frac{d}{dx}\cos x = -\sin x, \qquad \frac{d}{dx}\tan x = \sec^2 x

You can see the first one on the graphs. Where y=sin⁡xy = \sin x peaks, its slope is 00, and that is where cos⁡x=0\cos x = 0. Where sin⁡x\sin x crosses the axis going up, its slope is steepest, 11, and cos⁡x=1\cos x = 1 there.

The tangent rule comes from the quotient rule: tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}, so its derivative is cos⁡xcos⁡x−sin⁡x(−sin⁡x)cos⁡2x=1cos⁡2x=sec⁡2x\frac{\cos x \cos x - \sin x(-\sin x)}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x, using sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.

Radians only. These rules hold when xx is in radians. In degrees every derivative would carry an extra factor of π180\frac{\pi}{180}. Set your calculator to radians before evaluating.

With the chain rule. Multiply by the derivative of the angle: ddxsin⁡(kx)=kcos⁡(kx)\frac{d}{dx}\sin(kx) = k\cos(kx). For a power of a trig function, sin⁡2x\sin^2 x means (sin⁡x)2(\sin x)^2, so the outside function is the square: ddxsin⁡2x=2sin⁡xcos⁡x\frac{d}{dx}\sin^2 x = 2\sin x\cos x.

Sign check. The minus sign belongs to the derivative of cosine. Mixing that up is the most common error, so say the cycle to yourself: sine to cosine, cosine to negative sine.

Extreme values. Setting a trig derivative to zero gives a trig equation. Solve it on the interval the question gives, then compare the function at each solution and at the endpoints to find the absolute maximum and minimum.

Leave answers exact, with π\pi and roots, unless a decimal is asked for.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

  1. Easy

    Differentiate y=3sin⁡x−2cos⁡xy = 3\sin x - 2\cos x.
    Show solution
    1. ddx3sin⁡x=3cos⁡x\frac{d}{dx}3\sin x = 3\cos x and ddx(−2cos⁡x)=2sin⁡x\frac{d}{dx}(-2\cos x) = 2\sin x.

    Answer: y′=3cos⁡x+2sin⁡xy' = 3\cos x + 2\sin x

  2. Easy

    Differentiate y=sin⁡(4x)−cos⁡(2x)y = \sin(4x) - \cos(2x).
    Show solution
    1. Chain rule on each: ddxsin⁡(4x)=4cos⁡(4x)\frac{d}{dx}\sin(4x) = 4\cos(4x).
    2. ddx(−cos⁡(2x))=2sin⁡(2x)\frac{d}{dx}(-\cos(2x)) = 2\sin(2x).

    Answer: y′=4cos⁡(4x)+2sin⁡(2x)y' = 4\cos(4x) + 2\sin(2x)

  3. Medium

    Differentiate y=sin⁡2(3x)y = \sin^2(3x) and simplify.
    Show solution
    1. Rewrite as y=(sin⁡(3x))2y = (\sin(3x))^2. Outside: the square. Inside: sin⁡(3x)\sin(3x), with derivative 3cos⁡(3x)3\cos(3x).
    2. y′=2sin⁡(3x)⋅3cos⁡(3x)=6sin⁡(3x)cos⁡(3x)y' = 2\sin(3x) \cdot 3\cos(3x) = 6\sin(3x)\cos(3x).
    3. Double angle formula: 2sin⁡θcos⁡θ=sin⁡(2θ)2\sin\theta\cos\theta = \sin(2\theta), so y′=3sin⁡(6x)y' = 3\sin(6x).

    Answer: y′=6sin⁡(3x)cos⁡(3x)=3sin⁡(6x)y' = 6\sin(3x)\cos(3x) = 3\sin(6x)

  4. Medium

    Find the equation of the tangent line to y=tan⁡xy = \tan x at x=π4x = \frac{\pi}{4}.
    Show solution
    1. Point: tan⁡π4=1\tan\frac{\pi}{4} = 1.
    2. Slope: sec⁡2π4=1cos⁡2π4=112=2\sec^2\frac{\pi}{4} = \frac{1}{\cos^2\frac{\pi}{4}} = \frac{1}{\frac{1}{2}} = 2.
    3. y−1=2(x−π4)y - 1 = 2\left(x - \frac{\pi}{4}\right), so y=2x+1−π2y = 2x + 1 - \frac{\pi}{2}.

    Answer: y=2x+1−π2y = 2x + 1 - \frac{\pi}{2}

  5. Test level

    Find the absolute maximum and minimum of f(x)=x+2cos⁡xf(x) = x + 2\cos x on [0,π][0, \pi].
    Show solution
    1. f′(x)=1−2sin⁡x=0f'(x) = 1 - 2\sin x = 0 gives sin⁡x=12\sin x = \frac{1}{2}, so x=π6x = \frac{\pi}{6} or x=5π6x = \frac{5\pi}{6} in the interval.
    2. Endpoints: f(0)=2f(0) = 2 and f(π)=π−2≈1.14f(\pi) = \pi - 2 \approx 1.14.
    3. Critical numbers: f(π6)=π6+3≈2.26f\left(\frac{\pi}{6}\right) = \frac{\pi}{6} + \sqrt{3} \approx 2.26 and f(5π6)=5π6−3≈0.89f\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} - \sqrt{3} \approx 0.89.
    4. Compare all four values.

    Answer: Absolute max π6+3≈2.26\frac{\pi}{6} + \sqrt{3} \approx 2.26 at x=π6x = \frac{\pi}{6}; absolute min 5π6−3≈0.89\frac{5\pi}{6} - \sqrt{3} \approx 0.89 at x=5π6x = \frac{5\pi}{6}.

Learn It on Martian Lab

Martian Lab teaches this in the Derivatives of Sinusoidal Functions lesson of its MCV4U Derivatives of Trig, Exp and Log unit. Watch it, then take the Derivatives of Trig, Exp and Log practice tests to see if it stuck.

Watch the Derivatives of Sinusoidal Functions lesson free

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