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To approximate the value of \( \sqrt[3]{63} \) using differentials, we follow these steps:

### Step 1: Define the function
Let \( f(x) = \sqrt[3]{x} = x^{1/3} \), so we want to approximate \( f(63) \).

### Step 2: Choose a point close to 63
We'll choose \( x_0 = 64 \) because \( \sqrt[3]{64} = 4 \), which is an easy value to compute and close to 63.

### Step 3: Find the derivative
We need the derivative of \( f(x) = x^{1/3} \). Using the power rule:
\[
f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}
\]
At \( x_0 = 64 \), this becomes:
\[
f'(64) = \frac{1}{3\sqrt[3]{64^2}} = \frac{1}{3(64)^{2/3}} = \frac{1}{3 \times 16} = \frac{1}{48}
\]

### Step 4: Use the linear approximation formula
The differential approximation formula is:
\[
f(x) \approx f(x_0) + f'(x_0)(x - x_0)
\]
Substituting \( x_0 = 64 \), \( f(64) = 4 \), and \( f'(64) = \frac{1}{48} \), we get:
\[
f(63) \approx 4 + \frac{1}{48}(63 - 64)
\]
\[
f(63) \approx 4 + \frac{1}{48}(-1)
\]
\[
f(63) \approx 4 - \frac{1}{48} = 4 - 0.0208 = 3.9792
\]

### Conclusion:
Using differentials, the approximate value of \( \sqrt[3]{63} \) is \( 3.9792 \).

Would you like further details or have any questions?

### Related Questions:
1. What is the purpose of differentials in calculus approximations?
2. How can we apply differentials to approximate values for higher powers, such as \( \sqrt[4]{x} \)?
3. Why is \( x_0 = 64 \) a better choice than other numbers close to 63 for this problem?
4. What happens if we try to approximate \( \sqrt[3]{62} \) using differentials?
5. How does the error in a differential approximation compare to other approximation methods?

### Tip:
When using differentials for approximation, always choose a point \( x_0 \) near the target value where the function is easy to evaluate exactly.

Use differentials to approximate the value of the cube root of 63. Round your answer to four decimal places, if necessary.

Learn how to use differentials to approximate the value of the cube root of 63. This step-by-step guide covers key concepts such as derivatives and the linear approximation formula. Ideal for students in Grades 11-12 or early university calculus.

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