To solve for the derivative of $h(x) = \sin(\ln(x^5))$, we'll proceed with the following steps:

Step 1: Rewrite the Function

The function is given by: $h(x) = \sin(\ln(x^5))$ Using the logarithmic identity $\ln(x^5) = 5 \ln(x)$, we can rewrite $h(x)$ as: $h(x) = \sin(5 \ln(x))$

Step 2: Differentiate $h(x)$ with Respect to $x$

We want to find $h'(x)$, the derivative of $h(x)$.

This requires applying the chain rule. Let: $u = 5 \ln(x)$ so that $h(x) = \sin(u)$. The derivative $h'(x)$ then becomes: $h'(x) = \cos(u) \cdot u'$

Step 3: Find $u'$ with Respect to $x$

Since $u = 5 \ln(x)$, we find $u'$ by differentiating $5 \ln(x)$ with respect to $x$: $u' = 5 \cdot \frac{1}{x} = \frac{5}{x}$

Step 4: Substitute $u$ and $u'$ Back into $h'(x)$

Now, we can substitute $u = 5 \ln(x)$ and $u' = \frac{5}{x}$ into the derivative formula: $h'(x) = \cos(5 \ln(x)) \cdot \frac{5}{x}$

Final Answer

Thus, the derivative $h'(x)$ is: $h'(x) = \frac{5 \cos(5 \ln(x))}{x}$

Would you like more details on any step or have any questions?

Here are some related questions to expand your understanding:

1. How would the derivative change if the function was $h(x) = \sin(\ln(x^n))$ with any exponent $n$?
2. What if we replaced $\sin$ with another trigonometric function, like $\cos$?
3. How does the chain rule apply to composite functions like this one?
4. What other identities could be useful for simplifying functions with logarithmic and exponential expressions?
5. How would you integrate a function similar to $h(x)$ instead of differentiating it?

Tip: When working with logarithmic and trigonometric functions together, simplify expressions using logarithmic properties first before applying derivatives.