1. Differentiate $f(x) = x^2 + \frac{1}{x^2}$. - Rewrite $\frac{1}{x^2}$ as $x^{-2}$. - Use power rule: $\frac{d}{dx}x^n = nx^{n-1}$. - So, $f'(x) = 2x + (-2)x^{-3} = 2x - \frac{2}{x^3}$. 2. Differentiate $f(x) = (x^2 + x + 1)(x^2 + 2)$. - Use product rule: $(uv)' = u'v + uv'$. - Let $u = x^2 + x + 1$, $v = x^2 + 2$. - Compute derivatives: $u' = 2x + 1$, $v' = 2x$. - Thus, $f'(x) = (2x + 1)(x^2 + 2) + (x^2 + x + 1)(2x)$. - Expand: $= (2x^3 + 4x + x^2 + 2) + (2x^3 + 2x^2 + 2x)$. - Combine like terms: $4x^3 + 3x^2 + 6x + 2$. 3. Differentiate $f(x) = \frac{3x - 1}{2x + 1}$. - Use quotient rule: $\frac{u'v - uv'}{v^2}$. - Let $u = 3x - 1$, $v = 2x + 1$. - Derivatives: $u' = 3$, $v' = 2$. - So, $f'(x) = \frac{3(2x + 1) - (3x - 1)(2)}{(2x + 1)^2}$. - Numerator: $6x + 3 - 6x + 2 = 5$. - Hence, $f'(x) = \frac{5}{(2x + 1)^2}$. 4. Differentiate $f(x) = \frac{2x}{9 + x^2}$. - Use quotient rule with $u = 2x$, $v = 9 + x^2$. - Derivatives: $u' = 2$, $v' = 2x$. - So, $f'(x) = \frac{2(9 + x^2) - 2x(2x)}{(9 + x^2)^2} = \frac{18 + 2x^2 - 4x^2}{(9 + x^2)^2} = \frac{18 - 2x^2}{(9 + x^2)^2}$. 5. Differentiate $f(x) = \frac{1 - x}{2 + x}$. - Using quotient rule with $u = 1 - x$, $v = 2 + x$. - Derivatives: $u' = -1$, $v' = 1$. - So, $f'(x) = \frac{-1(2 + x) - (1 - x)(1)}{(2 + x)^2} = \frac{-2 - x - 1 + x}{(2 + x)^2} = \frac{-3}{(2 + x)^2}$. 6. Differentiate $f(x) = 6x^3 \cdot 7x^4$. - First simplify: $f(x) = 42x^{3 + 4} = 42x^7$. - Use power rule: $f'(x) = 42 \cdot 7 x^{6} = 294 x^{6}$. 7. Differentiate $f(x) = 3x^4 - 2x^3 + x^2 - 4x + 2$. - Differentiate each term: - $\frac{d}{dx}3x^4 = 12x^3$, $\frac{d}{dx}(-2x^3) = -6x^2$, $\frac{d}{dx}x^2 = 2x$, $\frac{d}{dx}(-4x) = -4$, derivative of constant 2 is 0. - Combine: $f'(x) = 12x^3 - 6x^2 + 2x - 4$. Final answers: 1. $2x - \frac{2}{x^3}$ 2. $4x^3 + 3x^2 + 6x + 2$ 3. $\frac{5}{(2x + 1)^2}$ 4. $\frac{18 - 2x^2}{(9 + x^2)^2}$ 5. $\frac{-3}{(2 + x)^2}$ 6. $294 x^6$ 7. $12x^3 - 6x^2 + 2x - 4$