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.