Questions: In Exercises 1-8, given y=f(u) and u=g(x), find dy / dx=f'(g(x)) g'(x). 2. y=2u^3, u=8x-1

Transcript text: In Exercises 1-8, given $y=f(u)$ and $u=g(x)$, find $d y / d x=f^{\prime}(g(x)) g^{\prime}(x)$. 2. $y=2 u^{3}, \quad u=8 x-1$

Solution

Solution Steps

To find $ \frac{dy}{dx} $ given $ y = f(u) $ and $ u = g(x) $, we use the chain rule. The chain rule states that $ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $. Here, $ y = 2u^3 $ and $ u = 8x - 1 $. First, compute $ \frac{dy}{du} $ by differentiating $ y $ with respect to $ u $. Then, compute $ \frac{du}{dx} $ by differentiating $ u $ with respect to $ x $. Finally, multiply these derivatives to find $ \frac{dy}{dx} $.

Step 1: Define the Functions

We are given the functions: \[ y = 2u^3 \] \[ u = 8x - 1 \]

Step 2: Compute the Derivative $ \frac{dy}{du} $

To find $ \frac{dy}{du} $, we differentiate $ y $ with respect to $ u $: \[ \frac{dy}{du} = \frac{d}{du}(2u^3) = 6u^2 \]

Step 3: Compute the Derivative $ \frac{du}{dx} $

Next, we differentiate $ u $ with respect to $ x $: \[ \frac{du}{dx} = \frac{d}{dx}(8x - 1) = 8 \]

Step 4: Apply the Chain Rule

Using the chain rule, we find $ \frac{dy}{dx} $: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 6u^2 \cdot 8 = 48u^2 \]

Step 5: Substitute $ u $ Back into the Expression

Now, we substitute $ u = 8x - 1 $ back into the expression for $ \frac{dy}{dx} $: \[ \frac{dy}{dx} = 48(8x - 1)^2 \]

Final Answer

Thus, the final expression for $ \frac{dy}{dx} $ is: \[ \boxed{\frac{dy}{dx} = 48(8x - 1)^2} \]

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