find the derivative of f(x)=-4/x^5 +sqrtx

To find the derivative of the function \( f(x) = -\frac{4}{x^5} + \sqrt{x} \), we need to apply the rules of differentiation.

1. **Rewrite the function**:
   - The first term \( -\frac{4}{x^5} \) can be written as \( -4x^{-5} \).
   - The second term \( \sqrt{x} \) is the same as \( x^{1/2} \).

Thus, the function becomes:
\[
f(x) = -4x^{-5} + x^{1/2}
\]

2. **Differentiate each term** using the power rule \( \frac{d}{dx} \left( x^n \right) = n x^{n-1} \).

   - For the first term \( -4x^{-5} \), we apply the power rule:
     \[
     \frac{d}{dx} (-4x^{-5}) = -4 \cdot (-5) x^{-5-1} = 20x^{-6}
     \]

   - For the second term \( x^{1/2} \), we apply the power rule:
     \[
     \frac{d}{dx} \left( x^{1/2} \right) = \frac{1}{2}x^{\frac{1}{2}-1} = \frac{1}{2}x^{-1/2}
     \]

3. **Combine the results**:
\[
f'(x) = 20x^{-6} + \frac{1}{2}x^{-1/2}
\]

4. **Simplify the expression**:
\[
f'(x) = \frac{20}{x^6} + \frac{1}{2\sqrt{x}}
\]

So, the derivative of \( f(x) = -\frac{4}{x^5} + \sqrt{x} \) is:
\[
f'(x) = \frac{20}{x^6} + \frac{1}{2\sqrt{x}}
\]

Would you like a detailed explanation on any part of this process?

Here are 5 questions to further expand:
1. How do we handle negative exponents when differentiating?
2. Can the chain rule be applied in this case?
3. What is the general form of the power rule for differentiation?
4. How do you differentiate more complex square root functions?
5. How does the derivative of \( x^n \) change with different values of \( n \)?

**Tip**: Always rewrite square roots and fractions involving powers of \(x\) in a simpler form before differentiating!

find the derivative of f(x)=-4/x^5 + sqrtx

Learn how to find the derivative of f(x) = -4/x^5 + sqrt(x) by applying the power rule in calculus. This step-by-step guide simplifies the function and explains each differentiation step, making it perfect for students in Grades 11-12 or early college-level calculus.

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