https://file.aiquickdraw.com/mathbot/file/17274251364786bmpnu71.jpg

Let's solve the derivative problem from the image step by step.

The given function is:
\[
f(x) = \frac{4x^8 - 5x^6 + 7x^4}{8x^4}
\]

### Step 1: Simplify the function
We can first simplify the given function by dividing each term in the numerator by \(8x^4\):
\[
f(x) = \frac{4x^8}{8x^4} - \frac{5x^6}{8x^4} + \frac{7x^4}{8x^4}
\]

Simplifying each term:
\[
f(x) = \frac{4}{8}x^{8-4} - \frac{5}{8}x^{6-4} + \frac{7}{8}x^{4-4}
\]

This simplifies to:
\[
f(x) = \frac{1}{2}x^4 - \frac{5}{8}x^2 + \frac{7}{8}
\]

### Step 2: Find the first derivative
Now, differentiate the simplified function term by term:
\[
f'(x) = \frac{d}{dx} \left( \frac{1}{2}x^4 - \frac{5}{8}x^2 + \frac{7}{8} \right)
\]

Applying the power rule for each term:

1. Derivative of \(\frac{1}{2}x^4\) is \( \frac{1}{2} \times 4x^{3} = 2x^3 \)
2. Derivative of \(- \frac{5}{8}x^2\) is \( - \frac{5}{8} \times 2x = - \frac{5}{4}x \)
3. Derivative of \(\frac{7}{8}\) is 0 since it's a constant.

So, the first derivative is:
\[
f'(x) = 2x^3 - \frac{5}{4}x
\]

### Final Answer:
The first derivative of the function is:
\[
f'(x) = 2x^3 - \frac{5}{4}x
\]

Would you like more details on any specific step?

Here are some related questions to explore:
1. How do you apply the power rule to differentiate functions?
2. What are the conditions when constants vanish during differentiation?
3. How do you simplify rational expressions before differentiation?
4. What is the significance of higher-order terms in a polynomial function?
5. How does the degree of a polynomial affect its behavior as \(x\) approaches infinity?

**Tip:** Simplifying a function before differentiating can help reduce errors and save time.

Find the first derivative of the following function: f(x) = (4x^8 - 5x^6 + 7x^4) / 8x^4

Learn how to find the first derivative of the function f(x) = (4x^8 - 5x^6 + 7x^4) / 8x^4. This solution involves simplification of rational expressions and application of the power rule in calculus. Suitable for high school students in Grades 11-12.

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