**Problem:** Find $\frac{dy}{dx}$ for each pair given $y=f(u)$ and $u=g(x)$ using the chain rule $\frac{dy}{dx} = f'(g(x))g'(x)$ for Exercises 1 to 8.
1. Given $y=6u-9$ and $u=\frac{1}{2}x^4$
Step 1: Differentiate $y$ with respect to $u$:
$$\frac{dy}{du} = 6$$
Step 2: Differentiate $u$ with respect to $x$:
$$\frac{du}{dx} = \frac{1}{2} \cdot 4x^3 = 2x^3$$
Step 3: Apply the chain rule:
$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 6 \cdot 2x^3 = 12x^3$$
2. Given $y=2u^3$ and $u=8x-1$
Step 1: Differentiate $y$ with respect to $u$:
$$\frac{dy}{du} = 6u^2$$
Step 2: Differentiate $u$ with respect to $x$:
$$\frac{du}{dx} = 8$$
Step 3: Apply the chain rule:
$$\frac{dy}{dx} = 6u^2 \cdot 8 = 48(8x -1)^2$$
3. Given $y=\sin u$ and $u=3x+1$
Step 1: Differentiate $y$ with respect to $u$:
$$\frac{dy}{du} = \cos u$$
Step 2: Differentiate $u$ with respect to $x$:
$$\frac{du}{dx} = 3$$
Step 3: Apply the chain rule:
$$\frac{dy}{dx} = \cos (3x+1) \cdot 3 = 3\cos (3x+1)$$
4. Given $y=\cos u$ and $u=e^{-x}$
Step 1: Differentiate $y$ with respect to $u$:
$$\frac{dy}{du} = -\sin u$$
Step 2: Differentiate $u$ with respect to $x$:
$$\frac{du}{dx} = -e^{-x}$$
Step 3: Apply the chain rule:
$$\frac{dy}{dx} = -\sin (e^{-x}) \cdot (-e^{-x}) = e^{-x} \sin (e^{-x})$$
5. Given $y=\sqrt{u}$ and $u=\sin x$
Step 1: Differentiate $y$ with respect to $u$:
$$\frac{dy}{du} = \frac{1}{2\sqrt{u}}$$
Step 2: Differentiate $u$ with respect to $x$:
$$\frac{du}{dx} = \cos x$$
Step 3: Apply the chain rule:
$$\frac{dy}{dx} = \frac{1}{2\sqrt{\sin x}} \cdot \cos x = \frac{\cos x}{2\sqrt{\sin x}}$$
6. Given $y=\sin u$ and $u=x - \cos x$
Step 1: Differentiate $y$ with respect to $u$:
$$\frac{dy}{du} = \cos u$$
Step 2: Differentiate $u$ with respect to $x$:
$$\frac{du}{dx} = 1 - (-\sin x) = 1 + \sin x$$
Step 3: Apply the chain rule:
$$\frac{dy}{dx} = \cos (x - \cos x) \cdot (1 + \sin x)$$
7. Given $y=\tan u$ and $u=\pi x^2$
Step 1: Differentiate $y$ with respect to $u$:
$$\frac{dy}{du} = \sec^2 u$$
Step 2: Differentiate $u$ with respect to $x$:
$$\frac{du}{dx} = 2\pi x$$
Step 3: Apply the chain rule:
$$\frac{dy}{dx} = \sec^2 (\pi x^2) \cdot 2\pi x = 2\pi x \sec^2 (\pi x^2)$$
8. Given $y=-\sec u$ and $u=\frac{1}{x} + 7x$
Step 1: Differentiate $y$ with respect to $u$:
$$\frac{dy}{du} = -\sec u \tan u$$
Step 2: Differentiate $u$ with respect to $x$:
$$\frac{du}{dx} = -\frac{1}{x^2} + 7 = 7 - \frac{1}{x^2}$$
Step 3: Apply the chain rule:
$$\frac{dy}{dx} = -\sec (\frac{1}{x} + 7x) \tan (\frac{1}{x} + 7x) \cdot \left(7 - \frac{1}{x^2}\right)$$
Chain Rule 1 8
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