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What are the derivatives of the following functions: 2) y = a + e + pi + cos 14^circ + 2x 4) y = 2x(4x2 + 2x − 6 + 2/x - 4/x2) 6) y = sin x + 2 cos x - 5 tan x?

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here are the derivatives for the ticked problems:

2) Given y=a+e+π+cos⁡14∘+2xy = a + e + \pi + \cos 14^\circ + 2x

  1. Identify constants.

    The terms aa, ee, π\pi, and cos⁡14∘\cos 14^\circ are all constants.
    The derivative of any constant is 0.

  2. Differentiate each term.

    dydx=ddx(a)+ddx(e)+ddx(π)+ddx(cos⁡14∘)+ddx(2x)\frac{dy}{dx} = \frac{d}{dx}(a) + \frac{d}{dx}(e) + \frac{d}{dx}(\pi) + \frac{d}{dx}(\cos 14^\circ) + \frac{d}{dx}(2x) dydx=0+0+0+0+2(1)x1−1\frac{dy}{dx} = 0 + 0 + 0 + 0 + 2(1)x^{1-1} dydx=2x0\frac{dy}{dx} = 2x^0 dydx=2\frac{dy}{dx} = 2

    The final answer is 2\boxed{2}.

    4) Given y=2x(4x2+2x−6+2x−4x2)y = 2x\left(4x^2 + 2x - 6 + \frac{2}{x} - \frac{4}{x^2}\right)

  3. Expand the expression by multiplying $2x$ by each term inside the parenthesis.

    y=2x(4x2)+2x(2x)−2x(6)+2x(2x)−2x(4x2)y = 2x(4x^2) + 2x(2x) - 2x(6) + 2x\left(\frac{2}{x}\right) - 2x\left(\frac{4}{x^2}\right) y=8x3+4x2−12x+4−8xy = 8x^3 + 4x^2 - 12x + 4 - \frac{8}{x}
  4. Rewrite the term with a negative exponent.

    y=8x3+4x2−12x+4−8x−1y = 8x^3 + 4x^2 - 12x + 4 - 8x^{-1}
  5. Differentiate each term using the power rule ddx(axn)=anxn−1\frac{d}{dx}(ax^n) = anx^{n-1}.

    dydx=8(3)x3−1+4(2)x2−1−12(1)x1−1+0−8(−1)x−1−1\frac{dy}{dx} = 8(3)x^{3-1} + 4(2)x^{2-1} - 12(1)x^{1-1} + 0 - 8(-1)x^{-1-1} dydx=24x2+8x−12x0+8x−2\frac{dy}{dx} = 24x^2 + 8x - 12x^0 + 8x^{-2} dydx=24x2+8x−12+8x2\frac{dy}{dx} = 24x^2 + 8x - 12 + \frac{8}{x^2}

    The final answer is 24x2+8x−12+8x2\boxed{24x^2 + 8x - 12 + \frac{8}{x^2}}.

    6) Given y=sin⁡x+2cos⁡x−5tan⁡xy = \sin x + 2 \cos x - 5 \tan x

  6. Differentiate each trigonometric term using standard derivative rules.

    Recall:
    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

  7. Apply the derivative rules to each term.

    dydx=ddx(sin⁡x)+2ddx(cos⁡x)−5ddx(tan⁡x)\frac{dy}{dx} = \frac{d}{dx}(\sin x) + 2\frac{d}{dx}(\cos x) - 5\frac{d}{dx}(\tan x) dydx=cos⁡x+2(−sin⁡x)−5(sec⁡2x)\frac{dy}{dx} = \cos x + 2(-\sin x) - 5(\sec^2 x) dydx=cos⁡x−2sin⁡x−5sec⁡2x\frac{dy}{dx} = \cos x - 2\sin x - 5\sec^2 x

    The final answer is cos⁡x−2sin⁡x−5sec⁡2x\boxed{\cos x - 2\sin x - 5\sec^2 x}.

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