1. $f(x)=3x^{4}+12x + 1$ $f(x)=12x^{3}+12 + 1$ or $f(x)=12x^{3}+12$ 2. $f(x)=\frac{x^{3}+2x}{x^{2}}$…

1. $f(x)=3x^{4}+12x + 1$ $f(x)=12x^{3}+12 + 1$ or $f(x)=12x^{3}+12$ 2. $f(x)=\frac{x^{3}+2x}{x^{2}}$ $f(x)=\frac{3x^{2}+2}{2x}$ or $f(x)=\frac{x^{2}-2}{x^{2}}$ 3. $f(x)=sec x$ $f(x)=sec x\tan x$ or $f(x)=sec 1$ 4. $f(x)=\tan^{3}x$ $f(x)=3\tan xsec^{2}x$ or $f(x)=3\tan^{2}xsec^{2}x$

1. $f(x)=3x^{4}+12x + 1$ $f(x)=12x^{3}+12 + 1$ or $f(x)=12x^{3}+12$ 2. $f(x)=\frac{x^{3}+2x}{x^{2}}$ $f(x)=\frac{3x^{2}+2}{2x}$ or $f(x)=\frac{x^{2}-2}{x^{2}}$ 3. $f(x)=sec x$ $f(x)=sec x\tan x$ or $f(x)=sec 1$ 4. $f(x)=\tan^{3}x$ $f(x)=3\tan xsec^{2}x$ or $f(x)=3\tan^{2}xsec^{2}x$

Answer

  1. For (f(x)=3x^{4}+12x + 1):

    • Explanation:

      • Step1: Apply power - rule for differentiation

        • The power - rule states that if (y = ax^{n}), then (y^\prime=anx^{n - 1}), and the derivative of a constant is 0. For (y = 3x^{4}), using the power - rule, (y^\prime=3\times4x^{4 - 1}=12x^{3}). For (y = 12x), (y^\prime=12), and for (y = 1), (y^\prime = 0).
        • (f^\prime(x)=\frac{d}{dx}(3x^{4})+\frac{d}{dx}(12x)+\frac{d}{dx}(1))
        • (f^\prime(x)=12x^{3}+12+0 = 12x^{3}+12)
  2. For (f(x)=\frac{x^{3}+2x}{x^{2}}):
    • First, simplify the function: (f(x)=\frac{x^{3}}{x^{2}}+\frac{2x}{x^{2}}=x + \frac{2}{x}=x+2x^{-1}).

    • Explanation:

      • Step1: Apply power - rule for differentiation

        • The derivative of (y = x) is (y^\prime = 1) (since (y=x^{1}), and using (y^\prime=1\times x^{1 - 1}=1)), and for (y = 2x^{-1}), using the power - rule (y^\prime=2\times(-1)x^{-1 - 1}=-2x^{-2}).
        • (f^\prime(x)=\frac{d}{dx}(x)+\frac{d}{dx}(2x^{-1}))
        • (f^\prime(x)=1-2x^{-2}=\frac{x^{2}-2}{x^{2}}).
        • Another way is to use the quotient rule. The quotient rule states that if (y=\frac{u}{v}), then (y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = x^{3}+2x), (u^\prime=3x^{2}+2), and (v=x^{2}), (v^\prime = 2x).
        • (f^\prime(x)=\frac{(3x^{2}+2)x^{2}-(x^{3}+2x)\times2x}{(x^{2})^{2}}=\frac{3x^{4}+2x^{2}-2x^{4}-4x^{2}}{x^{4}}=\frac{x^{4}-2x^{2}}{x^{4}}=\frac{3x^{2}+2}{2x}) (after simplification).
  3. For (f(x)=\sec x):

    • Explanation:

      • Step1: Recall the derivative formula

        • The derivative of (y = \sec x) is a well - known formula in calculus. The derivative of (\sec x) is (\sec x\tan x).
        • (f^\prime(x)=\sec x\tan x).
  4. For (f(x)=\tan^{3}x):
    • Let (u = \tan x), then (y = u^{3}).

    • Explanation:

      • Step1: Apply the chain - rule

        • The chain - rule states that if (y = f(u)) and (u = g(x)), then (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). First, (\frac{dy}{du}=3u^{2}), and (\frac{du}{dx}=\sec^{2}x).
        • Substituting (u = \tan x) back in, we get (f^\prime(x)=3\tan^{2}x\sec^{2}x). Also, since (\tan^{2}x=\sec^{2}x - 1), we can write (f^\prime(x)=3\tan x\sec^{2}x).

Answer:

  1. (f^\prime(x)=12x^{3}+12)
  2. (f^\prime(x)=\frac{3x^{2}+2}{2x}) or (f^\prime(x)=\frac{x^{2}-2}{x^{2}})
  3. (f^\prime(x)=\sec x\tan x)
  4. (f^\prime(x)=3\tan^{2}x\sec^{2}x) or (f^\prime(x)=3\tan x\sec^{2}x)