Answer
The derivative of \( y = x^{\frac{1}{3}} (3x^2 - 2 - 4x^{-3}) \) with respect to \( x \) is:
\[
y' = 7x^{\frac{4}{3}} - \frac{2}{3}x^{-\frac{2}{3}} + \frac{32}{3}x^{-\frac{11}{3}}.
\]
Solution
We start with
\[
y = x^{\frac{1}{3}} \left( 3x^2 - 2 - 4x^{-3} \right).
\]
Our goal is to differentiate \( y \) with respect to \( x \). We will use the product rule, which states that if
\[
y = u(x)v(x),
\]
then
\[
y' = u'(x)v(x) + u(x)v'(x).
\]
Let
\[
u(x) = x^{\frac{1}{3}} \quad \text{and} \quad v(x) = 3x^2 - 2 - 4x^{-3}.
\]
### Step 1. Differentiate \( u(x) \)
Using the power rule, the derivative of \( x^{\frac{1}{3}} \) is
\[
u'(x) = \frac{1}{3} x^{\frac{1}{3}-1} = \frac{1}{3} x^{-\frac{2}{3}}.
\]
### Step 2. Differentiate \( v(x) \)
Differentiate each term in \( v(x) \):
- For \( 3x^2 \):
\[
\frac{d}{dx}(3x^2) = 6x.
\]
- The derivative of the constant \(-2\) is \(0\).
- For \(-4x^{-3}\):
\[
\frac{d}{dx}(-4x^{-3}) = -4(-3)x^{-4} = 12x^{-4}.
\]
Thus,
\[
v'(x) = 6x + 12x^{-4}.
\]
### Step 3. Apply the Product Rule
Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the product rule:
\[
y' = u'(x)v(x) + u(x)v'(x) = \frac{1}{3} x^{-\frac{2}{3}} \left( 3x^2 - 2 - 4x^{-3} \right) + x^{\frac{1}{3}} \left( 6x + 12x^{-4} \right).
\]
### Step 4. Simplify Each Term
First, simplify the expression
\[
\frac{1}{3} x^{-\frac{2}{3}} \left( 3x^2 - 2 - 4x^{-3} \right)
\]
by distributing:
- For the term with \( 3x^2 \):
\[
\frac{1}{3} x^{-\frac{2}{3}} (3x^2) = x^{-\frac{2}{3}}x^2 = x^{\frac{4}{3}}.
\]
- For the term with \(-2\):
\[
\frac{1}{3} x^{-\frac{2}{3}} (-2) = -\frac{2}{3} x^{-\frac{2}{3}}.
\]
- For the term with \(-4x^{-3}\):
\[
\frac{1}{3} x^{-\frac{2}{3}}(-4x^{-3}) = -\frac{4}{3} x^{-\frac{2}{3}-3} = -\frac{4}{3} x^{-\frac{11}{3}}.
\]
Now, simplify the second part:
\[
x^{\frac{1}{3}} \left( 6x + 12x^{-4} \right)
\]
by distributing:
- For the term with \( 6x \):
\[
x^{\frac{1}{3}} (6x) = 6x^{\frac{1}{3}+1} = 6x^{\frac{4}{3}}.
\]
- For the term with \( 12x^{-4} \):
\[
x^{\frac{1}{3}} (12x^{-4}) = 12x^{\frac{1}{3}-4} = 12x^{-\frac{11}{3}}.
\]
### Step 5. Combine Like Terms
Now we combine the corresponding terms:
\[
y' = \left( x^{\frac{4}{3}} + 6x^{\frac{4}{3}} \right) - \frac{2}{3} x^{-\frac{2}{3}} + \left( -\frac{4}{3}x^{-\frac{11}{3}} + 12x^{-\frac{11}{3}} \right).
\]
Combine the \( x^{\frac{4}{3}} \) terms:
\[
x^{\frac{4}{3}} + 6x^{\frac{4}{3}} = 7x^{\frac{4}{3}}.
\]
Combine the \( x^{-\frac{11}{3}} \) terms:
\[
-\frac{4}{3} x^{-\frac{11}{3}} + 12x^{-\frac{11}{3}} = \left( -\frac{4}{3} + 12 \right)x^{-\frac{11}{3}} = \frac{-4 + 36}{3}x^{-\frac{11}{3}} = \frac{32}{3}x^{-\frac{11}{3}}.
\]
Thus, the derivative is
\[
y' = 7x^{\frac{4}{3}} - \frac{2}{3}x^{-\frac{2}{3}} + \frac{32}{3}x^{-\frac{11}{3}}.
\]
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