1. The problem has three parts. (a) Differentiate $2^{\cos^2 x}$ with respect to $\cos^2 x$. Step 1: Recognize that treating $2^{\cos^2 x}$ as an exponential function of $\cos^2 x$, the derivative with respect to $\cos^2 x$ uses the formula for exponential functions: $$\frac{d}{du} a^u = a^u \ln(a) \quad \text{where } a=2, u=\cos^2 x.$$ Step 2: Hence, $$\frac{d}{d(\cos^2 x)} 2^{\cos^2 x} = 2^{\cos^2 x} \ln 2.$$ (b) Given $$\tan^{-1}(x^2 + y^2) = a^2,$$ find $\frac{dy}{dx}$. Step 1: Differentiate both sides implicitly with respect to $x$: $$\frac{d}{dx} \tan^{-1}(x^2 + y^2) = \frac{d}{dx} a^2 = 0.$$ Step 2: Use the chain rule on the left: $$\frac{1}{1+(x^2+y^2)^2} \cdot \frac{d}{dx}(x^2 + y^2) = 0.$$ Step 3: Derivative of inside: $$2x + 2y \frac{dy}{dx}.$$ Step 4: Set equation: $$\frac{2x + 2y \frac{dy}{dx}}{1+(x^2 + y^2)^2} = 0.$$ Step 5: Multiply both sides by denominator and rearrange: $$2x + 2y \frac{dy}{dx} = 0 \implies 2y \frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}.$$ (c) Evaluate: $$\tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \right) \right].$$ Step 1: Evaluate inner $\cos^{-1}( \frac{\sqrt{3}}{2} )$. Recall $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ so $$\cos^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{6}.$$ Step 2: Calculate $2 \cdot \frac{\pi}{6} = \frac{\pi}{3}$. Step 3: Calculate $\sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}$. Step 4: Multiply by 2: $$2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}.$$ Step 5: So the expression inside $\tan^{-1}$ is $\sqrt{3}$. Step 6: Recall $\tan \frac{\pi}{3} = \sqrt{3}$, so $$\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}.$$ Final answers: (a) $$\frac{d}{d(\cos^2 x)} 2^{\cos^2 x} = 2^{\cos^2 x} \ln 2.$$ (b) $$\frac{dy}{dx} = -\frac{x}{y}.$$ (c) $$\tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \right) \right] = \frac{\pi}{3}.$$