What are the various rules involved when differentiating or integrating . How do we integrate or differentiate trigonometric functions like sinO

Mathematics

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Answer

−5sin⁡x−3cos⁡(3x)-5\sin x - 3\cos(3x)

−2cos⁡x+14sin⁡(4x)+C-2\cos x + \frac{1}{4}\sin(4x) + C

Step-by-step solution

6 steps

Okay, Ismaella. Let's look at the common rules for differentiation and integration, including trigonometric functions.

Differentiation Rules

Here are the main rules you'll use:

  1. Constant Rule: The derivative of a constant is zero.

    ddx(c)=0\frac{d}{dx}(c) = 0

    Example: ddx(5)=0\frac{d}{dx}(5) = 0

  2. Power Rule: (As discussed before) If y=xny = x^n, then dydx=nxn−1\frac{dy}{dx} = nx^{n-1}. Example: ddx(x3)=3x2\frac{d}{dx}(x^3) = 3x^2

  3. Constant Multiple Rule: If y=cf(x)y = cf(x), where cc is a constant, then dydx=cddx(f(x))\frac{dy}{dx} = c \frac{d}{dx}(f(x)). Example: ddx(4x2)=4ddx(x2)=4(2x)=8x\frac{d}{dx}(4x^2) = 4 \frac{d}{dx}(x^2) = 4(2x) = 8x

  4. Sum/Difference Rule: If y=f(x)±g(x)y = f(x) \pm g(x), then dydx=ddx(f(x))±ddx(g(x))\frac{dy}{dx} = \frac{d}{dx}(f(x)) \pm \frac{d}{dx}(g(x)). You differentiate each term separately. Example: ddx(x2+3x)=ddx(x2)+ddx(3x)=2x+3\frac{d}{dx}(x^2 + 3x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(3x) = 2x + 3

  5. Trigonometric Functions:

    • ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x
    • ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = -\sin x
    • ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x) = \sec^2 x
  6. Chain Rule: Used when you have a function inside another function, like y=f(g(x))y = f(g(x)).

    dydx=f′(g(x))⋅g′(x)\frac{dy}{dx} = f'(g(x)) \cdot g'(x)

    Example: If y=sin⁡(2x)y = \sin(2x), let u=2xu = 2x. Then y=sin⁡uy = \sin u. dydu=cos⁡u\frac{dy}{du} = \cos u and dudx=2\frac{du}{dx} = 2. So, dydx=cos⁡(2x)⋅2=2cos⁡(2x)\frac{dy}{dx} = \cos(2x) \cdot 2 = 2\cos(2x).

Integration Rules

Integration rules often mirror differentiation rules in reverse. Remember to add the constant of integration, CC.

  1. Constant Multiple Rule: ∫cf(x) dx=c∫f(x) dx\int c f(x)\,dx = c \int f(x)\,dx Example: ∫4x2 dx=4∫x2 dx=4(x33)+C=43x3+C\int 4x^2\,dx = 4 \int x^2\,dx = 4 \left(\frac{x^3}{3}\right) + C = \frac{4}{3}x^3 + C

  2. Sum/Difference Rule: ∫(f(x)±g(x)) dx=∫f(x) dx±∫g(x) dx\int (f(x) \pm g(x))\,dx = \int f(x)\,dx \pm \int g(x)\,dx. You integrate each term separately. Example: ∫(x2+3x) dx=∫x2 dx+∫3x dx=x33+3x22+C\int (x^2 + 3x)\,dx = \int x^2\,dx + \int 3x\,dx = \frac{x^3}{3} + \frac{3x^2}{2} + C

  3. Power Rule: (As discussed before) ∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C (for n≠−1n \neq -1). Example: ∫x3 dx=x44+C\int x^3\,dx = \frac{x^4}{4} + C

  4. Trigonometric Functions:

    • ∫cos⁡x dx=sin⁡x+C\int \cos x\,dx = \sin x + C
    • ∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + C
    • ∫sec⁡2x dx=tan⁡x+C\int \sec^2 x\,dx = \tan x + C

Examples with Trigonometric Functions

Example 1: Differentiate y=5cos⁡x−sin⁡(3x)y = 5\cos x - \sin(3x)

  1. Differentiate the first term, 5cos⁡x5\cos x.

    Using the Constant Multiple Rule and the derivative of cos⁡x\cos x:

    ddx(5cos⁡x)=5ddx(cos⁡x)=5(−sin⁡x)=−5sin⁡x\frac{d}{dx}(5\cos x) = 5 \frac{d}{dx}(\cos x) = 5(-\sin x) = -5\sin x
  2. Differentiate the second term, −sin⁡(3x)-\sin(3x).

    Using the Chain Rule: Let u=3xu = 3x, so dudx=3\frac{du}{dx} = 3.
    Then ddx(sin⁡(3x))=cos⁡(3x)⋅ddx(3x)=cos⁡(3x)⋅3=3cos⁡(3x)\frac{d}{dx}(\sin(3x)) = \cos(3x) \cdot \frac{d}{dx}(3x) = \cos(3x) \cdot 3 = 3\cos(3x).
    So, the derivative of −sin⁡(3x)-\sin(3x) is −3cos⁡(3x)-3\cos(3x).

  3. Combine the results using the Difference Rule.

    dydx=−5sin⁡x−3cos⁡(3x)\frac{dy}{dx} = -5\sin x - 3\cos(3x)

    The derivative is −5sin⁡x−3cos⁡(3x)\boxed{-5\sin x - 3\cos(3x)}.

    Example 2: Integrate ∫(2sin⁡x+cos⁡(4x)) dx\int (2\sin x + \cos(4x))\,dx

  4. Integrate the first term, 2sin⁡x2\sin x.

    Using the Constant Multiple Rule and the integral of sin⁡x\sin x:

    ∫2sin⁡x dx=2∫sin⁡x dx=2(−cos⁡x)=−2cos⁡x\int 2\sin x\,dx = 2 \int \sin x\,dx = 2(-\cos x) = -2\cos x
  5. Integrate the second term, cos⁡(4x)\cos(4x).

    This requires a substitution (a form of the reverse chain rule). Let u=4xu = 4x, so du=4 dxdu = 4\,dx, which means dx=14 dudx = \frac{1}{4}\,du.

    ∫cos⁡(4x) dx=∫cos⁡u(14 du)=14∫cos⁡u du=14sin⁡u+C2=14sin⁡(4x)+C2\int \cos(4x)\,dx = \int \cos u \left(\frac{1}{4}\,du\right) = \frac{1}{4} \int \cos u\,du = \frac{1}{4}\sin u + C_2 = \frac{1}{4}\sin(4x) + C_2
  6. Combine the results and the constant of integration.

    ∫(2sin⁡x+cos⁡(4x)) dx=−2cos⁡x+14sin⁡(4x)+C\int (2\sin x + \cos(4x))\,dx = -2\cos x + \frac{1}{4}\sin(4x) + C

    The integral is −2cos⁡x+14sin⁡(4x)+C\boxed{-2\cos x + \frac{1}{4}\sin(4x) + C}.

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