1. **Problem statement:** Find the derivatives of each function using the chain rule. 2. **Part (a):** $y=\sin^2(x^2)$ - Rewrite as $y=(\sin(x^2))^2$. - Outer function: $u^2$, inner function: $u=\sin(x^2)$. - Derivative: $\frac{dy}{dx} = 2\sin(x^2) \cdot \cos(x^2) \cdot 2x = 4x \sin(x^2) \cos(x^2)$. - Using double angle identity $\sin(2a) = 2\sin a \cos a$, this is $2x \sin(2x^2)$. 3. **Part (b):** $y=3^{\cos x}$ - Use $a^{f(x)} = e^{f(x) \ln a}$. - Derivative: $\frac{dy}{dx} = 3^{\cos x} \cdot \ln 3 \cdot (-\sin x) = -3^{\cos x} \ln 3 \sin x$. 4. **Part (c):** $y=\sqrt{\sin(3x) + 5}$ - Rewrite as $y=(\sin(3x)+5)^{1/2}$. - Derivative: $\frac{dy}{dx} = \frac{1}{2\sqrt{\sin(3x)+5}} \cdot \cos(3x) \cdot 3 = \frac{3 \cos(3x)}{2\sqrt{\sin(3x)+5}}$. 5. **Part (d):** $y=\cos\big((x^3 +3x^2 -8)^4\big)$ - Outer function: $\cos v$, inner: $v=(x^3 +3x^2 -8)^4$. - Derivative: $-\sin(v) \cdot 4(x^3 +3x^2 -8)^3 \cdot (3x^2 +6x) = -4(x^3 +3x^2 -8)^3 (3x^2 +6x) \sin((x^3 +3x^2 -8)^4)$. 6. **Part (e):** $y=\cos^4 (x^3 +3x^2 -8)$ - Rewrite as $y = (\cos(x^3 +3x^2 -8))^4$. - Outer function: $u^4$, inner: $u=\cos(x^3 +3x^2 -8)$. - Derivative: $4 \cos^3(x^3 +3x^2 -8) \cdot (-\sin(x^3 +3x^2 -8)) \cdot (3x^2 +6x) = -4 \cos^3(x^3 +3x^2 -8) \sin(x^3 +3x^2 -8) (3x^2 +6x)$. **Final answers:** (a) $\frac{dy}{dx} = 2x \sin(2x^2)$ (b) $\frac{dy}{dx} = -3^{\cos x} \ln 3 \sin x$ (c) $\frac{dy}{dx} = \frac{3 \cos(3x)}{2\sqrt{\sin(3x)+5}}$ (d) $\frac{dy}{dx} = -4(x^3 +3x^2 -8)^3 (3x^2 +6x) \sin((x^3 +3x^2 -8)^4)$ (e) $\frac{dy}{dx} = -4 \cos^3(x^3 +3x^2 -8) \sin(x^3 +3x^2 -8) (3x^2 +6x)$