I. Differentiate the following functions. Do NOT simplify once you have differentiated! [4 points each] \( \quad y=\arctan \left(\ln \left(\sin ^{2} x\right)\right) \) \( [2] \quad y=\left(4 x^{2}-2\right)^{10}\left(5 x^{3}-x\right)^{5} \) \( [3] \quad y=\frac{\sin \left(x^{3}-5 x^{2}\right)}{3 x^{2}-10} \) You must use the Quotient Rule on this problem.

To differentiate the given functions, let's go step-by-step: 1. For \( y=\arctan \left(\ln \left(\sin ^{2} x\right)\right) \): Using the chain rule, we first differentiate the outer function \(\arctan(u)\) where \(u = \ln(\sin^2 x)\). The derivative of \(\arctan(u)\) is \(\frac{1}{1+u^2}\), so: \[ \frac{dy}{dx} = \frac{1}{1+(\ln(\sin^2 x))^2} \cdot \frac{d}{dx}\left(\ln(\sin^2 x)\right) \] Next, we need \(\frac{d}{dx}\left(\ln(\sin^2 x)\right)\): \[ \frac{d}{dx}\left(\ln(\sin^2 x)\right) = \frac{1}{\sin^2 x} \cdot 2\sin x \cos x = \frac{2\sin x \cos x}{\sin^2 x} = 2\cot x \] Combining these results, we have: \[ \frac{dy}{dx} = \frac{2\cot x}{1+(\ln(\sin^2 x))^2} \] 2. For \( y=\left(4 x^{2}-2\right)^{10}\left(5 x^{3}-x\right)^{5} \): Using the product rule, we will differentiate it as follows: Let \( u = (4x^2 - 2)^{10} \) and \( v = (5x^3 - x)^{5} \). The product rule states that: \[ \frac{dy}{dx} = u'v + uv' \] Now we find \( u' \) using the chain rule: \[ u' = 10(4x^2 - 2)^{9} \cdot 8x = 80x(4x^2 - 2)^{9} \] Next, finding \( v' \): \[ v' = 5(5x^3 - x)^{4} \cdot (15x^2 - 1) \] Putting it all together, we have: \[ \frac{dy}{dx} = 80x(4x^2 - 2)^{9}(5x^3 - x)^{5} + (4x^2 - 2)^{10} \cdot 5(5x^3 - x)^{4}(15x^2 - 1) \] 3. For \( y=\frac{\sin \left(x^{3}-5 x^{2}\right)}{3 x^{2}-10} \): We will use the Quotient Rule. Let \( u = \sin(x^3 - 5x^2) \) and \( v = 3x^2 - 10 \). The Quotient Rule states: \[ \frac{dy}{dx} = \frac{u'v - uv'}{v^2} \] Now we differentiate \( u \) using the chain rule: \[ u' = \cos(x^3 - 5x^2) \cdot (3x^2 - 10) \quad \text{(derivative of the inner function)} \] Next, we differentiate \( v \): \[ v' = 6x \] Substituting \( u \), \( u' \), \( v \), and \( v' \) into the quotient rule formula, we have: \[ \frac{dy}{dx} = \frac{\cos(x^3 - 5x^2) \cdot (3x^2 - 10)(3x^2 - 10) - \sin(x^3 - 5x^2) \cdot 6x}{(3x^2 - 10)^2} \]