et g be the function given by g(x)=lim h->0 (cos(x+h)-cosx)/h.  what is the instantaneous rate of change of g with respect to x at x=pi/3?

The function \( g(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos(x)}{h} \) is the definition of the derivative of \( \cos(x) \). Therefore, \( g(x) \) is simply the derivative of \( \cos(x) \), which is \( -\sin(x) \).

So, we have:

\[
g(x) = -\sin(x)
\]

Now, we are asked to find the instantaneous rate of change of \( g \) with respect to \( x \) at \( x = \frac{\pi}{3} \). This means we need to compute \( g \left( \frac{\pi}{3} \right) \).

Substitute \( x = \frac{\pi}{3} \) into \( g(x) \):

\[
g \left( \frac{\pi}{3} \right) = -\sin \left( \frac{\pi}{3} \right)
\]

We know that \( \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \), so:

\[
g \left( \frac{\pi}{3} \right) = -\frac{\sqrt{3}}{2}
\]

Therefore, the instantaneous rate of change of \( g \) with respect to \( x \) at \( x = \frac{\pi}{3} \) is:

\[
\boxed{-\frac{\sqrt{3}}{2}}
\]

Would you like any more details or clarifications? Here are five related questions to consider:

1. How is the derivative of \( \cos(x) \) derived?
2. What is the geometric interpretation of the instantaneous rate of change?
3. How does the concept of a limit help in defining the derivative?
4. How can we compute the derivative of a trigonometric function like \( \sin(x) \)?
5. How would the answer change if we were asked for \( g'(x) \) instead?

**Tip**: The derivative of a function at a point gives the slope of the tangent line to the function at that point.

Let g be the function given by g(x)=lim h->0 (cos(x+h)-cosx)/h. What is the instantaneous rate of change of g with respect to x at x=pi/3?

Discover how to calculate the instantaneous rate of change of g(x) at x=pi/3 using the derivative of the cosine function. This solution explores key calculus concepts like limits, derivatives, and trigonometric functions, ideal for students in advanced high school or college calculus courses.

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