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To differentiate the function:

\[
f(x) = \sqrt{(x^2 - 2)^3 - 4}
\]

we'll proceed with the following steps:

### Step 1: Rewrite the Function
Rewrite the square root as a power of \( \frac{1}{2} \):
\[
f(x) = \left( (x^2 - 2)^3 - 4 \right)^{\frac{1}{2}}
\]

### Step 2: Apply the Chain Rule
To find \( f'(x) \), we apply the chain rule. Let:
\[
u = (x^2 - 2)^3 - 4
\]
so that:
\[
f(x) = u^{\frac{1}{2}}
\]

Using the chain rule:
\[
f'(x) = \frac{1}{2} u^{-\frac{1}{2}} \cdot u'
\]

### Step 3: Differentiate \( u \) with Respect to \( x \)
Now, let’s find \( u' \), where \( u = (x^2 - 2)^3 - 4 \).

1. Differentiate the first part \( (x^2 - 2)^3 \) using the chain rule:
   \[
   \frac{d}{dx} (x^2 - 2)^3 = 3(x^2 - 2)^2 \cdot 2x = 6x(x^2 - 2)^2
   \]

2. The derivative of \( -4 \) is 0.

So,
\[
u' = 6x(x^2 - 2)^2
\]

### Step 4: Substitute \( u \) and \( u' \) Back
Now substitute \( u = (x^2 - 2)^3 - 4 \) and \( u' = 6x(x^2 - 2)^2 \) into the expression for \( f'(x) \):
\[
f'(x) = \frac{1}{2} \left( (x^2 - 2)^3 - 4 \right)^{-\frac{1}{2}} \cdot 6x(x^2 - 2)^2
\]

### Final Answer
Thus, the derivative of \( f(x) \) is:
\[
f'(x) = \frac{3x(x^2 - 2)^2}{\sqrt{(x^2 - 2)^3 - 4}}
\]

Would you like a more detailed breakdown of any part of this process?

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Here are some related questions to further explore this topic:

1. How would the derivative change if the exponent in \((x^2 - 2)^3\) was different?
2. How do we handle the derivative if there’s an additional constant inside the square root?
3. What is the second derivative of this function?
4. How can we apply implicit differentiation to similar problems?
5. How would the derivative change if we were dealing with a cube root instead of a square root?

**Tip:** When differentiating functions involving powers and roots, always try rewriting them in terms of fractional exponents to make the application of the chain rule easier.

Learn how to find the derivative of the function f(x) = sqrt((x^2 - 2)^3 - 4) using the chain rule and power rule. This step-by-step solution covers advanced differentiation techniques suitable for Calculus I students.

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