let f(x)= 1/(7x^3) + 5/(9x^5). evaluate f'(3)

To find the derivative of the function \( f(x) = \frac{1}{7x^3} + \frac{5}{9x^5} \) and then evaluate it at \( x = 3 \), we'll follow these steps:

1. **Rewrite the function**: Express the terms with negative exponents for easier differentiation.
   \[
   f(x) = \frac{1}{7} x^{-3} + \frac{5}{9} x^{-5}
   \]

2. **Differentiate each term**: Apply the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \).
   \[
   \frac{d}{dx}\left( \frac{1}{7} x^{-3} \right) = \frac{1}{7} \cdot (-3) x^{-4} = -\frac{3}{7} x^{-4}
   \]
   \[
   \frac{d}{dx}\left( \frac{5}{9} x^{-5} \right) = \frac{5}{9} \cdot (-5) x^{-6} = -\frac{25}{9} x^{-6}
   \]

3. **Combine the derivatives**:
   \[
   f'(x) = -\frac{3}{7} x^{-4} - \frac{25}{9} x^{-6}
   \]

4. **Evaluate \( f'(x) \) at \( x = 3 \)**:
   \[
   f'(3) = -\frac{3}{7} (3)^{-4} - \frac{25}{9} (3)^{-6}
   \]
   Calculate each

Learn how to find the derivative of f(x) = 1/(7x^3) + 5/(9x^5) and evaluate f'(3). This problem involves differentiation using the power rule for negative exponents. Suitable for students in Grades 11-12.

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