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 \( f(x) \). Let's start with the given function:

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

Simplify the second term:

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

Now, let's differentiate \( f(x) \) with respect to \( x \):

The derivative of \( \frac{1}{7}x^3 \) is:

\[ 
\frac{d}{dx}\left(\frac{1}{7}x^3\right) = \frac{1}{7} \cdot 3x^{3-1} = \frac{3}{7}x^2 
\]

The derivative of \( x^5 \) is:

\[ 
\frac{d}{dx}(x^5) = 5x^{5-1} = 5x^4 
\]

Thus, the derivative \( f'(x) \) is:

\[ 
f'(x) = \frac{3}{7}x^2 + 5x^4 
\]

Now, we evaluate \( f'(x) \) at \( x = 3 \):

\[ 
f'(3) = \frac{3}{7}(3)^2 + 5(3)^4 
\]

Calculating each term:

\[
\frac{3}{7}(3)^2 = \frac{3}{7} \times 9 = \frac{27}{7} 
\]

\[
5(3)^4 = 5 \times 81 = 405 
\]

Adding these values:

\[
f'(3) = \frac{27}{7} + 405 
\]

Convert 405 to a fraction with a denominator of 7:

\[
405 = \frac{2835}{7} 
\]

So:

\[
f'(3) = \frac{27}{7} + \frac{2835}{7} = \frac{2862}{7} 
\]

Finally, calculate:

\[
\frac{2862}{7} = 408.8571 \approx 409 
\]

Thus, \( f'(3) = 409 \).

If you have any questions or need more details, feel free to ask! Here are some related questions that can help deepen your understanding:

1. What is the derivative of a function, and why is it important?
2. How do you apply the power rule for differentiation?
3. Can you explain why \( \frac{9}{9}x^5 \) simplifies to \( x^5 \)?
4. How would you find the second derivative \( f''(x) \) of the function?
5. What are some applications of derivatives in real-world problems?
6. How do you interpret the derivative value \( f'(3) \) in practical terms?
7. What is the geometric interpretation of the derivative at a point?
8. How would the derivative change if the function included a constant term?

**Tip:** Remember that simplifying expressions before differentiating can make the process easier and help avoid mistakes!

Learn how to evaluate the derivative f'(3) of the function f(x) = (1/7)x^3 + x^5 using the power rule for differentiation. This problem involves differentiating a polynomial function and calculating specific values. Suitable for advanced high school students and those studying calculus.

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