Derivatives Expressions F78834

1. **Problem statement:** Differentiate each given function. --- 2. **Part a:** $s = 2t^{-5} - \frac{t^3}{3} - 0.3t^{-2}$ - Use the power rule: $\frac{d}{dt} t^n = n t^{n-1}$. - Differentiate term by term: $$\frac{ds}{dt} = 2 \cdot (-5) t^{-6} - \frac{1}{3} \cdot 3 t^{2} - 0.3 \cdot (-2) t^{-3}$$ - Simplify coefficients: $$\frac{ds}{dt} = -10 t^{-6} - t^{2} + 0.6 t^{-3}$$ --- 3. **Part b:** $y = (5x^4 - 4x^5)(3x^2 + 2x^3)$ - Use the product rule: $\frac{d}{dx}(fg) = f' g + f g'$. - Let $f = 5x^4 - 4x^5$, $g = 3x^2 + 2x^3$. - Compute derivatives: $$f' = 20x^3 - 20x^4$$ $$g' = 6x + 6x^2$$ - Apply product rule: $$\frac{dy}{dx} = (20x^3 - 20x^4)(3x^2 + 2x^3) + (5x^4 - 4x^5)(6x + 6x^2)$$ - Expand and simplify: First term: $$20x^3 \cdot 3x^2 = 60x^5$$ $$20x^3 \cdot 2x^3 = 40x^6$$ $$-20x^4 \cdot 3x^2 = -60x^6$$ $$-20x^4 \cdot 2x^3 = -40x^7$$ Sum first term: $60x^5 + 40x^6 - 60x^6 - 40x^7 = 60x^5 - 20x^6 - 40x^7$ Second term: $$5x^4 \cdot 6x = 30x^5$$ $$5x^4 \cdot 6x^2 = 30x^6$$ $$-4x^5 \cdot 6x = -24x^6$$ $$-4x^5 \cdot 6x^2 = -24x^7$$ Sum second term: $30x^5 + 30x^6 - 24x^6 - 24x^7 = 30x^5 + 6x^6 - 24x^7$ - Add both terms: $$\frac{dy}{dx} = (60x^5 - 20x^6 - 40x^7) + (30x^5 + 6x^6 - 24x^7) = 90x^5 - 14x^6 - 64x^7$$ --- 4. **Part c:** $u = \frac{x^3 - 2x}{x^2 + x + 3}$ - Use the quotient rule: $\frac{d}{dx} \left( \frac{f}{g} \right) = \frac{f' g - f g'}{g^2}$. - Let $f = x^3 - 2x$, $g = x^2 + x + 3$. - Compute derivatives: $$f' = 3x^2 - 2$$ $$g' = 2x + 1$$ - Apply quotient rule: $$\frac{du}{dx} = \frac{(3x^2 - 2)(x^2 + x + 3) - (x^3 - 2x)(2x + 1)}{(x^2 + x + 3)^2}$$ - Expand numerator: First part: $$ (3x^2 - 2)(x^2 + x + 3) = 3x^4 + 3x^3 + 9x^2 - 2x^2 - 2x - 6 = 3x^4 + 3x^3 + 7x^2 - 2x - 6$$ Second part: $$ (x^3 - 2x)(2x + 1) = 2x^4 + x^3 - 4x^2 - 2x$$ - Subtract second part from first: $$3x^4 + 3x^3 + 7x^2 - 2x - 6 - (2x^4 + x^3 - 4x^2 - 2x) = (3x^4 - 2x^4) + (3x^3 - x^3) + (7x^2 + 4x^2) + (-2x + 2x) - 6 = x^4 + 2x^3 + 11x^2 - 6$$ - Final derivative: $$\frac{du}{dx} = \frac{x^4 + 2x^3 + 11x^2 - 6}{(x^2 + x + 3)^2}$$ --- 5. **Part d:** $y = \sqrt[3]{\frac{1}{1+x^2}} = \left(1+x^2\right)^{-\frac{1}{3}}$ - Use chain rule and power rule. - Let $f(x) = 1 + x^2$, then $y = f(x)^{-1/3}$. - Derivative: $$\frac{dy}{dx} = -\frac{1}{3} f(x)^{-4/3} \cdot 2x = -\frac{2x}{3 (1+x^2)^{4/3}}$$ --- 6. **Part f:** $h(x) = \frac{x}{2+x}$ - First derivative using quotient rule: $$h'(x) = \frac{(1)(2+x) - x(1)}{(2+x)^2} = \frac{2 + x - x}{(2+x)^2} = \frac{2}{(2+x)^2}$$ - Second derivative: $$h''(x) = \frac{d}{dx} \left( 2 (2+x)^{-2} \right) = 2 \cdot (-2)(2+x)^{-3} \cdot 1 = -\frac{4}{(2+x)^3}$$ --- **Final answers:** - a) $\frac{ds}{dt} = -10 t^{-6} - t^{2} + 0.6 t^{-3}$ - b) $\frac{dy}{dx} = 90x^5 - 14x^6 - 64x^7$ - c) $\frac{du}{dx} = \frac{x^4 + 2x^3 + 11x^2 - 6}{(x^2 + x + 3)^2}$ - d) $\frac{dy}{dx} = -\frac{2x}{3 (1+x^2)^{4/3}}$ - f) $h'(x) = \frac{2}{(2+x)^2}$ and $h''(x) = -\frac{4}{(2+x)^3}$