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.