`f(x)=(2x^(2)+3)^(5/3)(x+5)^((-1)/3)`

Answer

To differentiate the function \( f(x) = (2x^2 + 3)^{5/3} (x + 5)^{-1/3} \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative \( f'(x) \) is given by:

\[ f'(x) = u'v + uv' \]

### Step 1: Identify \( u \) and \( v \)

Let:
- \( u = (2x^2 + 3)^{5/3} \)
- \( v = (x + 5)^{-1/3} \)

### Step 2: Differentiate \( u \)

To differentiate \( u \), we will use the chain rule. The chain rule states that if \( y = g(h(x)) \), then \( \frac{dy}{dx} = g'(h(x)) \cdot h'(x) \).

1. Differentiate \( u \):
\[ u' = \frac{d}{dx}[(2x^2 + 3)^{5/3}] = \frac{5}{3}(2x^2 + 3)^{\frac{5}{3} - 1} \cdot (4x) = \frac{20x}{3}(2x^2 + 3)^{2/3} \]

### Step 3: Differentiate \( v \)

Now, differentiate \( v \):
\[ v' = \frac{d}{dx}[(x + 5)^{-1/3}] = -\frac{1}{3}(x + 5)^{-4/3} \cdot (1) = -\frac{1}{3}(x + 5)^{-4/3} \]

### Step 4: Apply the Product Rule

Now, we can apply the product rule:
\[ f'(x) = u'v + uv' \]
Substituting \( u, u', v, \) and \( v' \):
\[ f'(x) = \left(\frac{20x}{3}(2x^2 + 3)^{2/3}\right)(x + 5)^{-1/3} + (2x^2 + 3)^{5/3}\left(-\frac{1}{3}(x + 5)^{-4/3}\right) \]

### Step 5: Simplify the Expression

Now, let's simplify the expression:
\[ f'(x) = \frac{20x}{3}(2x^2 + 3)^{2/3}(x + 5)^{-1/3} - \frac{1}{3}(2x^2 + 3)^{5/3}(x + 5)^{-4/3} \]

### Step 6: Combine the Terms

To combine the terms, we can factor out common terms:
\[ f'(x) = \frac{1}{3}(2x^2 + 3)^{2/3}(x + 5)^{-4/3} \left(20x(x + 5) - (2x^2 + 3)\right) \]

### Step 7: Final Expression

Thus, the final derivative is:
\[ f'(x) = \frac{1}{3}(2x^2 + 3)^{2/3}(x + 5)^{-4/3} \left(20x^2 + 100x - 2x^2 - 3\right) \]
\[ = \frac{1}{3}(2x^2 + 3)^{2/3}(x + 5)^{-4/3} \left(18x^2 + 100x - 3\right) \]