Question

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 $f(x)=3\tan^{2}xsec^{2}x$

Explanation:

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 \(y = 1\), \(y^\prime = 0\).
• \(f^\prime(x)=\frac{d}{dx}(3x^{4})+\frac{d}{dx}(12x)+\frac{d}{dx}(1)\)
• \(f^\prime(x)=12x^{3}+12+0 = 12x^{3}+12\)

1. 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).

1. 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 \(\sec x\tan x\).
• \(f^\prime(x)=\sec x\tan x\).

1. 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 \(f^\prime(x)=3\tan x\sec^{2}x\)