The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of  on the interval 

Then take the difference of the two and divide by the interval.

Now find the derivative of the function; this will be solved for the value(s) found above.

Trigonometric derivative: 

Product rule: 

Using a calculator, we find the solution , which fits within the interval , satisfying the mean value theorem.

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of  on the interval 

Then take the difference of the two and divide by the interval.

Now find the derivative of the function; this will be solved for the value(s) found above.

Derivative of an exponential: 

Trigonometric derivative: 

Product rule: 

Using a calculator, we find the solution , which fits within the interval , satisfying the mean value theorem.

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of  on the interval 

Then take the difference of the two and divide by the interval.

Now find the derivative of the function; this will be solved for the value(s) found above.

Trigonometric derivative: 

Product rule: 

Using a calculator, we find the solutions , which fit within the interval , satisfying the mean value theorem.

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval  for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of  on the interval 

Then take the difference of the two and divide by the interval.

Now find the derivative of the function; this will be solved for the value(s) found above.

Derivative of an exponential: 

Trigonometric derivative: 

Using a calculator, there are multiple solutions; on the interval , the one that satisfies the mean value theorem is 

In order to find the velocity of a given point, you must first find the derivative of the position function.

In this case, the derivative of the position function is:

Then, find  when .

The answer is: 

We can solve this problem using the chain rule:

Let    and   

We want to find 

and

So, 

.

Multiplying the polynomials together and taking the derivative of each term will yield the correct function for y'.

Using the power rule for each term:

The power rule states,

.

Therefore the derivative becomes,

.

We could also find the answer through the product rule:

Notice that the product rule in this case is not more efficient, since you still have to FOIL out h(x)g'(x), and then group like terms together. It is important to recognize when using the product rule is an asset and when it is more complicated than necessary.

To find the answer, first find the first derivative, f''(x).

To find the derivative use the power rule which states,

.

Thus we get the following.

Plugging in 2 for x, we find:

When finding the derivative of functions combined with addition or subtraction, we can simply combine the functions with the appropriate operation (subtraction in this case) and take the derivative of each term in the resulting function.

 When we take the derivative of the difference we get:

When differentiating a function, you must understand how to find the derivative of different types of functions:

If  (constant), then .

If, then.

If , then .

There are multiple different rules that apply when differentiating different kinds of functions. 

In this case, you find the derivative of each portion of the function:

And so the simplified answer is: