Let’s calculate the derivatives of the given functions step by step.

1. $f(x) = 7x^7 + 8x - 9$

The derivative is: [ f'(x) = \frac{d}{dx}[7x^7] + \frac{d}{dx}[8x] - \frac{d}{dx}[9] = 49x^6 + 8. ]

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2. $f(x) = \frac{7}{x^3} - \frac{x^4}{5} - 6\sqrt[3]{x}$

Rewriting the terms for differentiation: $f(x) = 7x^{-3} - \frac{1}{5}x^4 - 6x^{\frac{1}{3}}$ The derivative is: [ f'(x) = \frac{d}{dx}[7x^{-3}] - \frac{d}{dx}\left[\frac{1}{5}x^4\right] - \frac{d}{dx}[6x^{\frac{1}{3}}] ] $f'(x) = -21x^{-4} - \frac{4}{5}x^3 - 2x^{-\frac{2}{3}}.$

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3. $f(x) = (3x - 1)\sin x$

Using the product rule $f'(x) = u'v + uv'$, where $u = 3x - 1$ and $v = \sin x$: $f'(x) = \frac{d}{dx}(3x - 1)\cdot \sin x + (3x - 1)\cdot \frac{d}{dx}(\sin x)$ $f'(x) = 3\sin x + (3x - 1)\cos x.$

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4. $f(x) = \frac{2 - x}{\cos x}$

Using the quotient rule $f'(x) = \frac{u'v - uv'}{v^2}$, where $u = 2 - x$ and $v = \cos x$: $u' = -1, \quad v' = -\sin x$ $f'(x) = \frac{(-1)(\cos x) - (2 - x)(-\sin x)}{(\cos x)^2}$ $f'(x) = \frac{-\cos x + (2 - x)\sin x}{\cos^2 x}.$

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5. $f(x) = (5x^3 + 4)(x - 2)$

Using the product rule $f'(x) = u'v + uv'$, where $u = 5x^3 + 4$ and $v = x - 2$: $u' = 15x^2, \quad v' = 1$ $f'(x) = (15x^2)(x - 2) + (5x^3 + 4)(1)$ $f'(x) = 15x^3 - 30x^2 + 5x^3 + 4 = 20x^3 - 30x^2 + 4.$

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6. $f(x) = (x^3 - 2)^4$

Using the chain rule $f'(x) = n[g(x)]^{n-1}g'(x)$, where $g(x) = x^3 - 2$ and $n = 4$: $f'(x) = 4(x^3 - 2)^3 \cdot \frac{d}{dx}(x^3 - 2)$ $f'(x) = 4(x^3 - 2)^3 \cdot 3x^2 = 12x^2(x^3 - 2)^3.$

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7. $f(x) = \cos\left(4x + \frac{5\pi}{3}\right)$

Using the chain rule $f'(x) = -\sin[g(x)]g'(x)$, where $g(x) = 4x + \frac{5\pi}{3}$: $f'(x) = -\sin\left(4x + \frac{5\pi}{3}\right) \cdot \frac{d}{dx}(4x + \frac{5\pi}{3})$ $f'(x) = -\sin\left(4x + \frac{5\pi}{3}\right) \cdot 4 = -4\sin\left(4x + \frac{5\pi}{3}\right).$

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Would you like more details on any step or have additional questions?

Related Questions:

1. Can you explain how to apply the chain rule in more detail?
2. How is the quotient rule derived and why does it work?
3. Could you provide examples of the product rule in different contexts?
4. What is the significance of higher-order derivatives in these functions?
5. How would these derivatives be used in real-world applications?

Tip:

Always simplify the function first, if possible, to make differentiation easier and reduce the risk of errors.