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

To evaluate \( f'(3) \), we first need to find the derivative of the function \( f(x) \). The function given is:

\[
f(x) = \frac{1}{7x^3} + \frac{9}{9x^5} 
\]

First, simplify the second term:

\[
f(x) = \frac{1}{7x^3} + \frac{1}{x^5} 
\]

Now let's rewrite these terms using negative exponents:

\[
f(x) = \frac{1}{7}x^{-3} + x^{-5}
\]

To differentiate \( f(x) \), we'll use the power rule for differentiation, which states that \(\frac{d}{dx}(x^n) = nx^{n-1}\).

### Differentiating \( f(x) \)

1. **Differentiate \( \frac{1}{7}x^{-3} \):**

   \[
   \frac{d}{dx}\left(\frac{1}{7}x^{-3}\right) = \frac{1}{7} \cdot (-3)x^{-3-1} = -\frac{3}{7}x^{-4}
   \]

2. **Differentiate \( x^{-5} \):**

   \[
   \frac{d}{dx}(x^{-5}) = -5x^{-5-1} = -5x^{-6}
   \]

Combining these, the derivative \( f'(x) \) is:

\[
f'(x) = -\frac{3}{7}x^{-4} - 5x^{-6}
\]

### Evaluating \( f'(3) \)

Substitute \( x = 3 \) into \( f'(x) \):

\[
f'(3) = -\frac{3}{7}(3)^{-4} - 5(3)^{-6}
\]

Calculate each term:

\[
-\frac{3}{7}(3)^{-4} = -\frac{3}{7} \cdot \frac{1}{3^4} = -\frac{3}{7} \cdot \frac{1}{81} = -\frac{3}{567} = -\frac{1}{189}
\]

\[
-5(3)^{-6} = -5 \cdot \frac{1}{3^6} = -5 \cdot \frac{1}{729} = -\frac{5}{729}
\]

Now, add these fractions:

\[
f'(3) = -\frac{1}{

Learn how to evaluate the derivative f'(3) for the function f(x) = 1/(7x^3) + 9/(9x^5) using differentiation and power rule. This problem involves differentiating a function with negative exponents and is suitable for advanced math learners.

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