Subjects calculus

Derivatives Functions Aa333F

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Question: User: g(x) = x3 + 7x + 6 h(x) = 7 + \frac{1}{x^2} f(x) = (x3 + 7x + 6)\left(7 + \frac{1}{x^2}\right) Find the derivative of each function. g'(x) = 3x2+7 h'(x) = -2x^{-3} f'(x) =
1. **State the problem:** Find the derivatives of the functions $$g(x) = x^3 + 7x + 6$$ $$h(x) = 7 + \frac{1}{x^2}$$ $$f(x) = (x^3 + 7x + 6)\left(7 + \frac{1}{x^2}\right)$$ 2. **Derivative of** $g(x)$: Use the power rule: $$\frac{d}{dx} x^n = nx^{n-1}$$ $$g'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(7x) + \frac{d}{dx}(6)$$ $$= 3x^{2} + 7 + 0 = 3x^{2} + 7$$ 3. **Derivative of** $h(x)$: Rewrite $$h(x) = 7 + x^{-2}$$ Derivative: $$h'(x) = 0 + \frac{d}{dx} x^{-2} = -2x^{-3}$$ 4. **Derivative of** $f(x)$: Use the product rule: $$f(x) = u(x) v(x)$$ where $$u(x) = x^3 + 7x + 6$$ and $$v(x) = 7 + x^{-2}$$ Product rule formula: $$f'(x) = u'(x) v(x) + u(x) v'(x)$$ Calculate derivatives: $$u'(x) = 3x^{2} + 7$$ $$v'(x) = 0 - 2x^{-3} = -2x^{-3}$$ Substitute: $$f'(x) = (3x^{2} + 7)(7 + x^{-2}) + (x^{3} + 7x + 6)(-2x^{-3})$$ 5. **Simplify**: Expand first term: $$ (3x^{2} + 7)(7 + x^{-2}) = 3x^{2} \cdot 7 + 3x^{2} \cdot x^{-2} + 7 \cdot 7 + 7 \cdot x^{-2} $$ $$= 21x^{2} + 3 + 49 + 7x^{-2} = 21x^{2} + 52 + 7x^{-2}$$ Expand second term: $$ (x^{3} + 7x + 6)(-2x^{-3}) = -2x^{3} x^{-3} - 14x x^{-3} - 12 x^{-3} $$ $$= -2 - 14 x^{-2} - 12 x^{-3}$$ 6. **Combine all terms:** $$f'(x) = (21x^{2} + 52 + 7x^{-2}) + (-2 - 14 x^{-2} - 12 x^{-3})$$ $$= 21x^{2} + (52 - 2) + (7x^{-2} - 14 x^{-2}) - 12 x^{-3}$$ $$= 21x^{2} + 50 - 7 x^{-2} - 12 x^{-3}$$ **Final answer:** $$\boxed{f'(x) = 21x^{2} + 50 - 7 x^{-2} - 12 x^{-3}}$$ This completes the derivatives for all three functions.