(9) To find the derivative of the function $$ f(x) $$, I'll first rewrite it from the image: $$ f(x)=\frac{\left(x^{-\frac{1}{2}}\right)\left(x^{3}-343\right)(x+7)}{3\left(x^{2}-x+49\right)} $$ This is a product and quotient rule problem. Let's solve this step by step.

Question

UpStudy AI · Accepted Answer

Certainly! Let's find the derivative of the function step by step.

Given:
$$
f(x) = \frac{\left(x^{-\frac{1}{2}}\right)\left(x^{3}-343\right)(x+7)}{3\left(x^{2}-x+49\right)}
$$

To find $$ f'(x) $$, we'll use the **Quotient Rule** combined with the **Product Rule**. Here's a detailed breakdown:

## Step 1: Simplify the Function (If Possible)

Before delving into differentiation, it's often helpful to simplify the function to make the differentiation process easier.

1. **Factor $$ x^3 - 343 $$:**
   $$
   x^3 - 343 = x^3 - 7^3 = (x - 7)(x^2 + 7x + 49)
   $$
   
2. **Express $$ f(x) $$ with the factored numerator:**
   $$
   f(x) = \frac{x^{-1/2} \cdot (x - 7)(x^2 + 7x + 49)(x + 7)}{3(x^2 - x + 49)}
   $$

Observing the denominator $$ x^2 - x + 49 $$, it doesn’t factor out with any part of the numerator. So, we'll proceed without further simplification.

## Step 2: Identify Components for the Quotient Rule

The **Quotient Rule** states that if $$ f(x) = \frac{N(x)}{D(x)} $$, then:
$$
f'(x) = \frac{N'(x)D(x) - N(x)D'(x)}{[D(x)]^2}
$$

Let's define:
- **Numerator ($$ N(x) $$):**
  $$
  N(x) = x^{-1/2} \cdot (x^3 - 343) \cdot (x + 7)
  $$
  
- **Denominator ($$ D(x) $$):**
  $$
  D(x) = 3(x^2 - x + 49)
  $$

## Step 3: Compute $$ D'(x) $$

Differentiate $$ D(x) $$ with respect to $$ x $$:
$$
D(x) = 3(x^2 - x + 49)
$$
$$
D'(x) = 3(2x - 1) = 6x - 3
$$

## Step 4: Compute $$ N'(x) $$ Using the Product Rule

$$ N(x) $$ is a product of three functions:
1. $$ u(x) = x^{-1/2} $$
2. $$ v(x) = x^3 - 343 $$
3. $$ w(x) = x + 7 $$

The **Product Rule** for three functions states:
$$
N'(x) = u'(x)v(x)w(x) + u(x)v'(x)w(x) + u(x)v(x)w'(x)
$$

Let's compute each derivative:

1. **Derivative of $$ u(x) $$:**
   $$
   u(x) = x^{-1/2} \Rightarrow u'(x) = -\frac{1}{2}x^{-3/2}
   $$
   
2. **Derivative of $$ v(x) $$:**
   $$
   v(x) = x^3 - 343 \Rightarrow v'(x) = 3x^2
   $$
   
3. **Derivative of $$ w(x) $$:**
   $$
   w(x) = x + 7 \Rightarrow w'(x) = 1
   $$

Now, plug these into the Product Rule:
$$
N'(x) = \left(-\frac{1}{2}x^{-3/2}\right)(x^3 - 343)(x + 7) + x^{-1/2}(3x^2)(x + 7) + x^{-1/2}(x^3 - 343)(1)
$$

Simplify term by term:

1. **First Term:**
   $$
   -\frac{1}{2}x^{-3/2}(x^3 - 343)(x + 7)
   $$
   
2. **Second Term:**
   $$
   3x^{-1/2}x^2(x + 7) = 3x^{3/2}(x + 7)
   $$
   
3. **Third Term:**
   $$
   x^{-1/2}(x^3 - 343) = x^{5/2} - 343x^{-1/2}
   $$

Thus, 
$$
N'(x) = -\frac{1}{2}x^{-3/2}(x^3 - 343)(x + 7) + 3x^{3/2}(x + 7) + x^{5/2} - 343x^{-1/2}
$$

## Step 5: Apply the Quotient Rule

Now, apply the **Quotient Rule**:
$$
f'(x) = \frac{N'(x)D(x) - N(x)D'(x)}{[D(x)]^2}
$$

Substitute $$ N(x) $$, $$ N'(x) $$, $$ D(x) $$, and $$ D'(x) $$ into the formula:
$$
f'(x) = \frac{\left(-\frac{1}{2}x^{-3/2}(x^3 - 343)(x + 7) + 3x^{3/2}(x + 7) + x^{5/2} - 343x^{-1/2}\right) \cdot 3(x^2 - x + 49) - x^{-1/2}(x^3 - 343)(x + 7) \cdot (6x - 3)}{\left[3(x^2 - x + 49)\right]^2}
$$

## Step 6: Simplify the Expression

This expression can be quite cumbersome, but we can look for simplifications:

1. **Factor out common terms:** Notice that $$ x^{-1/2} $$ appears in multiple terms.
2. **Combine like terms:** Group terms with the same exponents.
3. **Simplify polynomials:** Expand and combine polynomial expressions where possible.

**However**, given the complexity, it's often beneficial to leave the derivative in its factored form unless further simplification is required for a specific purpose.

## Final Expression for $$ f'(x) $$

After applying the Quotient and Product Rules, the derivative $$ f'(x) $$ is:

$$
f'(x) = \frac{-\frac{1}{2}x^{-3/2}(x^3 - 343)(x + 7) + 3x^{3/2}(x + 7) + x^{5/2} - 343x^{-1/2}}{3(x^2 - x + 49)} - \frac{x^{-1/2}(x^3 - 343)(x + 7)(6x - 3)}{[3(x^2 - x + 49)]^2}
$$

**Note:** This expression can be further simplified based on the requirements or by factoring out common terms. However, due to its complexity, it's often practical to handle specific evaluations or simplifications case by case.

---

If you need further simplification or have any questions about specific steps, feel free to ask!
To find the derivative of $$ f(x) = \frac{x^{-1/2}(x^3 - 343)(x + 7)}{3(x^2 - x + 49)} $$, use the Quotient and Product Rules. After differentiation, the derivative $$ f'(x) $$ is: $$ f'(x) = \frac{-\frac{1}{2}x^{-3/2}(x^3 - 343)(x + 7) + 3x^{3/2}(x + 7) + x^{5/2} - 343x^{-1/2}}{3(x^2 - x + 49)} - \frac{x^{-1/2}(x^3 - 343)(x + 7)(6x - 3)}{[3(x^2 - x + 49)]^2} $$ This expression can be further simplified by factoring out common terms or expanding polynomials as needed.

(9) To find the derivative of the function \( f(x) \), I'll first rewrite it from the image: \[ f(x)=\frac{\left(x^{-\frac{1}{2}}\right)\left(x^{3}-343\right)(x+7)}{3\left(x^{2}-x+49\right)} \] This is a product and quotient rule problem. Let's solve this step by step.

Real Tutor Solution

Answer

Solution

Mind Expander

Related Questions

Latest Calculus Questions