In the following find the approximate values, using differentials :
`(32.15)^(1//5)`

Answer

To find the approximate value of \( (32.15)^{1/5} \) using differentials, we can follow these steps:

### Step 1: Identify a nearby value
We start by identifying a value close to \( 32.15 \) for which we can easily calculate the fifth root. The number \( 32 \) is a good choice because \( 32^{1/5} = 2 \).

### Step 2: Define the variables
Let:
- \( x = 32 \)
- \( \Delta x = 32.15 - 32 = 0.15 \)

### Step 3: Define the function
We define the function \( y = x^{1/5} \).

### Step 4: Calculate \( y \) at \( x = 32 \)
We calculate:
\[ y = 32^{1/5} = 2 \]

### Step 5: Find the derivative \( \frac{dy}{dx} \)
Next, we need to find the derivative of \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{5} x^{-4/5} \]

### Step 6: Evaluate the derivative at \( x = 32 \)
Now, we evaluate \( \frac{dy}{dx} \) at \( x = 32 \):
\[ \frac{dy}{dx} = \frac{1}{5} (32)^{-4/5} \]
Calculating \( 32^{-4/5} \):
\[ 32^{1/5} = 2 \quad \text{(since \( 32 = 2^5 \))} \]
Thus,
\[ 32^{-4/5} = \frac{1}{2^4} = \frac{1}{16} \]
Now substituting this back:
\[ \frac{dy}{dx} = \frac{1}{5} \cdot \frac{1}{16} = \frac{1}{80} \]

### Step 7: Calculate \( \Delta y \)
Using the formula for differentials:
\[ \Delta y \approx \frac{dy}{dx} \cdot \Delta x \]
Substituting the values we have:
\[ \Delta y \approx \frac{1}{80} \cdot 0.15 = \frac{0.15}{80} = 0.001875 \]

### Step 8: Find the approximate value of \( (32.15)^{1/5} \)
Now, we can find the approximate value:
\[ (32.15)^{1/5} \approx y + \Delta y = 2 + 0.001875 = 2.001875 \]

### Final Answer
Thus, the approximate value of \( (32.15)^{1/5} \) is:
\[ \boxed{2.001875} \]