To find the derivative of $f(x) = \cos(g(x))$, we’ll use the chain rule.

Solution:

Given: $f(x) = \cos(g(x))$

1. Differentiate the outer function $\cos(g(x))$ with respect to $g(x)$: $\frac{d}{dg} \cos(g(x)) = -\sin(g(x))$

2. Multiply by the derivative of the inner function $g(x)$ with respect to $x$: $f'(x) = -\sin(g(x)) \cdot g'(x)$

Final Answer:

$f'(x) = -\sin(g(x)) \cdot g'(x)$

Would you like further clarification on any step?

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Here are some related questions to deepen your understanding:

1. How would the derivative change if $f(x) = \sin(g(x))$?
2. What if $f(x) = \cos(x^2 + 3x)$? How would you find the derivative?
3. How does the chain rule apply when differentiating $f(x) = \cos(g(h(x)))$?
4. What is the second derivative of $f(x) = \cos(g(x))$?
5. How would you handle the derivative if $g(x)$ were a trigonometric function like $\sin(x)$ or $\tan(x)$?

Tip: The chain rule is essential for differentiating compositions of functions, especially in trigonometric, exponential, and logarithmic contexts.