To find \(\frac{dy}{dx}\) when \(y = 5(\ln(x))^{-4}\), we can proceed by using the chain rule.

### Solution
1. Rewrite \(y\):
   \[
   y = 5(\ln(x))^{-4} = \frac{5}{(\ln(x))^4}
   \]

2. Differentiate with respect to \(x\):
   We can use the chain rule here. Let \(u = \ln(x)\), so \(y = 5u^{-4}\). Then, differentiate \(y\) with respect to \(u\) and \(u\) with respect to \(x\).

   - First, find \(\frac{dy}{du}\):
     \[
     \frac{dy}{du} = 5 \cdot (-4) u^{-5} = -20 u^{-5} = -\frac{20}{(\ln(x))^5}
     \]

   - Next, find \(\frac{du}{dx}\):
     \[
     \frac{du}{dx} = \frac{d}{dx}(\ln(x)) = \frac{1}{x}
     \]

3. Combine using the chain rule:
   \[
   \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{20}{(\ln(x))^5} \cdot \frac{1}{x} = -\frac{20}{x(\ln(x))^5}
   \]

### Final Answer
\[
\frac{dy}{dx} = -\frac{20}{x(\ln(x))^5}
\]

Would you like a deeper explanation of any of the steps?

Here are some related questions for further practice:

1. What is \(\frac{dy}{dx}\) if \(y = (\ln(x))^{-3}\)?
2. How would the derivative change if \(y = \ln(x)^{-2} + x^2\)?
3. What is the general form of the derivative of \(y = c(\ln(x))^n\), where \(c\) and \(n\) are constants?
4. How would you differentiate \(y = \frac{1}{(\ln(x))^2}\)?
5. If \(y = \sin((\ln(x))^{-1})\), what is \(\frac{dy}{dx}\)?

**Tip:** When differentiating expressions involving powers of functions, rewriting them in simpler forms (like fractions) often helps simplify the differentiation process.

Find the derivative of y=5(ln(x))^-4 with respect to x using the chain rule. This solution covers key calculus concepts, including logarithmic differentiation and applying the chain rule for functions of powers. Suitable for students in advanced high school calculus or introductory college calculus.

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