The differentiation of function `r(x)= (3x + 2)^(3/2)`w.r.t x is

Options

A. `9/2(3x + 2)^(1/2)`
B. `3/2(3x + 2)^(1/2)`
C. `3/9(3x + 2)^(-1/2)`
D. `9/2(3x+ 2)^(-3/2)`

Answer

To differentiate the function \( r(x) = (3x + 2)^{\frac{3}{2}} \) with respect to \( x \), we will use the chain rule. Here’s a step-by-step solution:

### Step 1: Identify the outer and inner functions
We can identify the outer function and the inner function:
- Outer function: \( u^{\frac{3}{2}} \), where \( u = 3x + 2 \)
- Inner function: \( u = 3x + 2 \)

### Step 2: Apply the chain rule
The chain rule states that if you have a composite function \( r(x) = f(g(x)) \), then the derivative is given by:
\[ \frac{dr}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \]
In our case, we need to find \( \frac{dr}{du} \) and \( \frac{du}{dx} \).

### Step 3: Differentiate the outer function
Differentiate the outer function \( u^{\frac{3}{2}} \):
\[ \frac{dr}{du} = \frac{3}{2} u^{\frac{3}{2} - 1} = \frac{3}{2} u^{\frac{1}{2}} \]

### Step 4: Differentiate the inner function
Now differentiate the inner function \( u = 3x + 2 \):
\[ \frac{du}{dx} = 3 \]

### Step 5: Combine using the chain rule
Now we can combine the derivatives using the chain rule:
\[ \frac{dr}{dx} = \frac{dr}{du} \cdot \frac{du}{dx} = \left(\frac{3}{2} u^{\frac{1}{2}}\right) \cdot 3 \]

### Step 6: Substitute back the inner function
Substituting \( u = 3x + 2 \) back into the equation:
\[ \frac{dr}{dx} = \frac{3}{2} (3x + 2)^{\frac{1}{2}} \cdot 3 \]
\[ = \frac{9}{2} (3x + 2)^{\frac{1}{2}} \]

### Final Answer
Thus, the differentiation of the function \( r(x) = (3x + 2)^{\frac{3}{2}} \) with respect to \( x \) is:
\[ \frac{dr}{dx} = \frac{9}{2} (3x + 2)^{\frac{1}{2}} \]

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