1. Given $y = 6u - 9$ and $u = \frac{1}{2}x^4$, find $\frac{dy}{dx}$. 1. State the problem: We want to find the derivative of $y$ with respect to $x$. 2. Use the chain rule formula: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ 3. Calculate $\frac{dy}{du}$: Since $y = 6u - 9$, then $$\frac{dy}{du} = 6$$ 4. Calculate $\frac{du}{dx}$: Since $u = \frac{1}{2}x^4$, then $$\frac{du}{dx} = \frac{1}{2} \cdot 4x^3 = 2x^3$$ 5. Multiply the derivatives: $$\frac{dy}{dx} = 6 \cdot 2x^3 = 12x^3$$ 3. Given $y = \sin u$ and $u = 3x + 1$, find $\frac{dy}{dx}$. 1. State the problem: Find the derivative of $y$ with respect to $x$. 2. Use the chain rule: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ 3. Calculate $\frac{dy}{du}$: Since $y = \sin u$, $$\frac{dy}{du} = \cos u$$ 4. Calculate $\frac{du}{dx}$: Since $u = 3x + 1$, $$\frac{du}{dx} = 3$$ 5. Multiply the derivatives: $$\frac{dy}{dx} = \cos(3x + 1) \cdot 3 = 3\cos(3x + 1)$$ 5. Given $y = x^2 \sin^4 x + x \cos^{-2} x$, find $\frac{dy}{dx}$. 1. State the problem: Find the derivative of $y$ with respect to $x$. 2. Use the sum rule: $$\frac{dy}{dx} = \frac{d}{dx}(x^2 \sin^4 x) + \frac{d}{dx}(x \cos^{-2} x)$$ 3. Differentiate the first term using product and chain rules: - Let $f = x^2$, $g = \sin^4 x$. Then $$\frac{d}{dx}(f g) = f' g + f g'$$ - Calculate $f' = 2x$. - Calculate $g' = 4 \sin^3 x \cdot \cos x$ (chain rule on $\sin^4 x$). - So, $$\frac{d}{dx}(x^2 \sin^4 x) = 2x \sin^4 x + x^2 \cdot 4 \sin^3 x \cos x = 2x \sin^4 x + 4x^2 \sin^3 x \cos x$$ 4. Differentiate the second term using product and chain rules: - Let $f = x$, $g = \cos^{-2} x = (\cos x)^{-2}$. - Calculate $f' = 1$. - Calculate $g' = -2 (\cos x)^{-3} \cdot (-\sin x) = 2 \sin x (\cos x)^{-3}$ (chain rule). - So, $$\frac{d}{dx}(x \cos^{-2} x) = 1 \cdot \cos^{-2} x + x \cdot 2 \sin x (\cos x)^{-3} = \cos^{-2} x + 2x \sin x (\cos x)^{-3}$$ 5. Combine the results: $$\frac{dy}{dx} = 2x \sin^4 x + 4x^2 \sin^3 x \cos x + \cos^{-2} x + 2x \sin x (\cos x)^{-3}$$