1. We are asked to find \( \frac{d y}{d x} \) for \( y = \cos \left( \frac{1 + x^{2}}{1 - x^{2}} \right) \). Step 1: Let \( u = \frac{1 + x^{2}}{1 - x^{2}} \). Step 2: Differentiate \( u \) using the quotient rule: $$ u' = \frac{(2x)(1 - x^{2}) - (1 + x^{2})(-2x)}{(1 - x^{2})^{2}} = \frac{2x(1 - x^{2}) + 2x(1 + x^{2})}{(1 - x^{2})^{2}} = \frac{2x(1 - x^{2} + 1 + x^{2})}{(1 - x^{2})^{2}} = \frac{4x}{(1 - x^{2})^{2}} $$ Step 3: Differentiate \( y \) using the chain rule: $$ \frac{d y}{d x} = -\sin(u) \cdot u' = -\sin \left( \frac{1 + x^{2}}{1 - x^{2}} \right) \cdot \frac{4x}{(1 - x^{2})^{2}} $$ 2. For \( y = \tan \left[ \sqrt{3 x^{2}} + \ln (5 x^{4}) \right] \), find \( \frac{d y}{d x} \). Step 1: Let \( v = \sqrt{3 x^{2}} + \ln (5 x^{4}) \). Step 2: Simplify inside \( v \): $$ \sqrt{3 x^{2}} = \sqrt{3} |x|, \quad \ln (5 x^{4}) = \ln 5 + 4 \ln x $$ Step 3: Derivative of the first term (assuming \( x>0 \) for derivative of \( |x|=x \)): $$ \frac{d}{d x} \sqrt{3} x = \sqrt{3} $$ Step 4: Derivative of the second term: $$ \frac{d}{d x} (\ln 5 + 4 \ln x) = 0 + \frac{4}{x} = \frac{4}{x} $$ Step 5: So, $$ v' = \sqrt{3} + \frac{4}{x} $$ Step 6: Using chain rule for \( y = \tan(v) \): $$ \frac{d y}{d x} = \sec^{2}(v) \cdot v' = \sec^{2} \left[ \sqrt{3 x^{2}} + \ln (5 x^{4}) \right] \left( \sqrt{3} + \frac{4}{x} \right) $$ 3. For \( y = (2 - 3x)^{\frac{1}{2}} \), find \( \frac{d y}{d x} \) using logarithmic differentiation. Step 1: Take natural logarithm both sides: $$ \ln y = \frac{1}{2} \ln (2 - 3x) $$ Step 2: Differentiate implicitly with respect to \( x \): $$ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \cdot \frac{-3}{2 - 3x} = \frac{-3}{2(2 - 3x)} $$ Step 3: Multiply both sides by \( y \): $$ \frac{dy}{dx} = y \cdot \frac{-3}{2 (2 - 3x)} = (2 - 3x)^{\frac{1}{2}} \cdot \frac{-3}{2 (2 - 3x)} $$ This can simplify to $$ \frac{dy}{dx} = -\frac{3}{2} (2 - 3x)^{-\frac{1}{2}} $$ # Final answers: \( \frac{d y}{d x} = -\sin \left( \frac{1 + x^{2}}{1 - x^{2}} \right) \cdot \frac{4x}{(1 - x^{2})^{2}} \) \( \frac{d y}{d x} = \sec^{2} \left[ \sqrt{3 x^{2}} + \ln (5 x^{4}) \right] \left( \sqrt{3} + \frac{4}{x} \right) \) \( \frac{d y}{d x} = -\frac{3}{2} (2 - 3x)^{-\frac{1}{2}} \)