In the following find the approximate values, using differentials :
`(33)^(-1//5)`

Answer

To find the approximate value of \( (33)^{-1/5} \) using differentials, we can follow these steps:

### Step 1: Define the function
Let \( y = f(x) = x^{-1/5} \).

### Step 2: Choose a point close to 33
We will choose a point \( x = 32 \) since it is close to 33 and easier to calculate.

### Step 3: Calculate \( f(32) \)
Now, we calculate \( f(32) \):
\[ f(32) = 32^{-1/5} \]
To find \( 32^{-1/5} \), we can recognize that \( 32 = 2^5 \):
\[ f(32) = (2^5)^{-1/5} = 2^{-1} = \frac{1}{2} = 0.5 \]

### Step 4: Find the derivative \( f'(x) \)
Next, we need to find the derivative \( f'(x) \):
\[ f'(x) = -\frac{1}{5} x^{-6/5} \]

### Step 5: Calculate \( f'(32) \)
Now we substitute \( x = 32 \) into the derivative:
\[ f'(32) = -\frac{1}{5} \cdot 32^{-6/5} \]
Since \( 32 = 2^5 \), we have:
\[ 32^{-6/5} = (2^5)^{-6/5} = 2^{-6} = \frac{1}{64} \]
Thus,
\[ f'(32) = -\frac{1}{5} \cdot \frac{1}{64} = -\frac{1}{320} \]

### Step 6: Calculate \( dy \)
Now, we can use the differential to find \( dy \):
\[ dy = f'(32) \cdot dx \]
Here, \( dx = 1 \) (since we are moving from 32 to 33):
\[ dy = -\frac{1}{320} \cdot 1 = -\frac{1}{320} \]

### Step 7: Approximate \( f(33) \)
Now we can approximate \( f(33) \):
\[ f(33) \approx f(32) + dy = 0.5 - \frac{1}{320} \]
To compute this, we need a common denominator:
\[ 0.5 = \frac{160}{320} \]
Thus,
\[ f(33) \approx \frac{160}{320} - \frac{1}{320} = \frac{159}{320} \]

### Step 8: Final result
Calculating \( \frac{159}{320} \) gives us approximately:
\[ \frac{159}{320} \approx 0.496875 \]

Therefore, the approximate value of \( (33)^{-1/5} \) is \( \approx 0.496875 \).

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