Finding the Instantaneous Rate of Change of a Function | Math Problem Solution with Derivatives at x = π/3

Let g be the function given by g(x)=limh→0 sin(x+h)−sinx / h. What is the instantaneous rate of change of g with respect to x at x=π/3?

To find the instantaneous rate of change of the function g at x = π/3, we need to evaluate the derivative of the function at that particular x-value

To find the instantaneous rate of change of the function g at x = π/3, we need to evaluate the derivative of the function at that particular x-value.

Given that g(x) = limh→0 [sin(x + h) – sin(x)] / h, we can start by simplifying the expression inside the limit:

sin(x + h) – sin(x) = 2sin( (x + h – x) / 2) * cos((x + h + x) / 2)
= 2sin(h/2) * cos((2x + h)/2)
= 2sin(h/2) * cos(x + h/2)

Therefore, g(x) = limh→0 [2sin(h/2) * cos(x + h/2)] / h

Now, let’s evaluate the limit as h approaches 0:

limh→0 [2sin(h/2) * cos(x + h/2)] / h

Using the fact that sin(x) / x approaches 1 as x approaches 0, we can rewrite the expression above as:

2 * sin(h/2) / (h/2) * cos(x + h/2)

As h approaches 0, the term sin(h/2) / (h/2) approaches 1, so the expression becomes:

2 * cos(x + h/2)

Now, let’s evaluate the expression at x = π/3:

g(π/3) = 2 * cos(π/3 + h/2)

Since we want to find the instantaneous rate of change, we take the derivative of g(x) with respect to x:

g'(x) = -2sin(x + h/2)

Substituting x = π/3:

g'(π/3) = -2sin(π/3 + h/2)

Finally, we can evaluate g'(π/3) by plugging in the value of x:

g'(π/3) = -2sin(π/3 + 0/2) = -2 sin(π/3) = -2 * (√3/2) = -√3

Therefore, the instantaneous rate of change of g with respect to x at x = π/3 is -√3.

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