1. **Find** $\frac{dy}{dx}$ for $$ y = \frac{x}{(x^6 - 2)^3} \cdot \left( \frac{x^2 - 3}{\sqrt{x+1}} \right)^{\frac{4}{3}} \left(2 + \frac{3}{x}\right)^7 \sqrt{x^2 + 2 + x^2} $$ Step 1: Simplify the last square root term: $$ \sqrt{x^2 + 2 + x^2} = \sqrt{2x^2 + 2} = \sqrt{2(x^2 + 1)} = \sqrt{2} \sqrt{x^2 + 1} $$ Step 2: Let $$ u = \frac{x}{(x^6 - 2)^3}, \quad v = \left( \frac{x^2 - 3}{\sqrt{x+1}} \right)^{\frac{4}{3}}, \quad w = \left(2 + \frac{3}{x}\right)^7, \quad z = \sqrt{2} \sqrt{x^2 + 1} $$ Then $$ y = u \cdot v \cdot w \cdot z $$ Step 3: Use the product rule for four functions: $$ \frac{dy}{dx} = u' v w z + u v' w z + u v w' z + u v w z' $$ Step 4: Compute each derivative: - For $u = \frac{x}{(x^6 - 2)^3}$, write as $$ u = x (x^6 - 2)^{-3} $$ Using product rule: $$ u' = 1 \cdot (x^6 - 2)^{-3} + x \cdot (-3)(x^6 - 2)^{-4} \cdot 6x^5 = (x^6 - 2)^{-3} - 18 x^6 (x^6 - 2)^{-4} $$ - For $v = \left( \frac{x^2 - 3}{(x+1)^{1/2}} \right)^{4/3}$, let $$ g = \frac{x^2 - 3}{(x+1)^{1/2}} $$ Then $$ v = g^{4/3} $$ Derivative: $$ v' = \frac{4}{3} g^{1/3} g' $$ Compute $g'$: $$ g = (x^2 - 3)(x+1)^{-1/2} $$ $$ g' = 2x (x+1)^{-1/2} + (x^2 - 3) \cdot \left(-\frac{1}{2}\right)(x+1)^{-3/2} = 2x (x+1)^{-1/2} - \frac{x^2 - 3}{2} (x+1)^{-3/2} $$ - For $w = \left(2 + \frac{3}{x}\right)^7$, let $$ h = 2 + \frac{3}{x} $$ Derivative: $$ w' = 7 h^6 \cdot h' = 7 \left(2 + \frac{3}{x}\right)^6 \cdot \left(-\frac{3}{x^2}\right) = -\frac{21}{x^2} \left(2 + \frac{3}{x}\right)^6 $$ - For $z = \sqrt{2} \sqrt{x^2 + 1} = \sqrt{2} (x^2 + 1)^{1/2}$, Derivative: $$ z' = \sqrt{2} \cdot \frac{1}{2} (x^2 + 1)^{-1/2} \cdot 2x = \sqrt{2} x (x^2 + 1)^{-1/2} $$ Step 5: Substitute all derivatives back into the product rule formula for $\frac{dy}{dx}$. --- 2. **Find** $\frac{d^2 y}{dx^2}$ for $$ y = x^7 - \frac{2}{x^3} + x^{3/2} + \frac{1}{\sqrt{x}} $$ Step 1: Compute first derivative $y'$: $$ y' = 7x^6 + 6x^{-4} + \frac{3}{2} x^{1/2} - \frac{1}{2} x^{-3/2} $$ Step 2: Compute second derivative $y''$: $$ y'' = 42 x^5 - 24 x^{-5} + \frac{3}{4} x^{-1/2} + \frac{3}{4} x^{-5/2} $$ --- 3. **Show** that if $$ y = \frac{x}{\sqrt{x^2 + 1}} $$ then $$ y' = \frac{1}{(x^2 + 1)^{3/2}} $$ Step 1: Write $y$ as $$ y = x (x^2 + 1)^{-1/2} $$ Step 2: Use product rule: $$ y' = 1 \cdot (x^2 + 1)^{-1/2} + x \cdot \left(-\frac{1}{2}\right)(x^2 + 1)^{-3/2} \cdot 2x = (x^2 + 1)^{-1/2} - x^2 (x^2 + 1)^{-3/2} $$ Step 3: Factor out $(x^2 + 1)^{-3/2}$: $$ y' = (x^2 + 1)^{-3/2} \left( (x^2 + 1) - x^2 \right) = (x^2 + 1)^{-3/2} \cdot 1 = \frac{1}{(x^2 + 1)^{3/2}} $$ --- 4. **Show** that if $$ y = x + \sqrt{x^2 + 1} $$ then $$ y' \sqrt{x^2 + 1} = y $$ Step 1: Compute $y'$: $$ y' = 1 + \frac{1}{2} (x^2 + 1)^{-1/2} \cdot 2x = 1 + \frac{x}{\sqrt{x^2 + 1}} $$ Step 2: Multiply both sides by $\sqrt{x^2 + 1}$: $$ y' \sqrt{x^2 + 1} = \sqrt{x^2 + 1} + x $$ Step 3: Recognize right side equals $y$: $$ y' \sqrt{x^2 + 1} = y $$ **Final answers:** 1. $\frac{dy}{dx} = u' v w z + u v' w z + u v w' z + u v w z'$ with derivatives as above. 2. $\frac{d^2 y}{dx^2} = 42 x^5 - 24 x^{-5} + \frac{3}{4} x^{-1/2} + \frac{3}{4} x^{-5/2}$. 3. $y' = \frac{1}{(x^2 + 1)^{3/2}}$. 4. $y' \sqrt{x^2 + 1} = y$.