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.

There is only one value within the open interval  that satisfies the mean value theorem, but there is still a value regardless.

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

Taking the derivative of the function at 

The slope of the tangent is

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

Taking the derivative of the function  at 

The slope of the tangent is

A line can be written as a function of the form 

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

Now to find this next constant, the normal line and the orignal function should intercept at 

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.

Of these solutions,  satisfies the mean value theorem by falling within 

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of an exponential: 

Trigonometric derivative: 

Taking the derivative of the function  at 

The slope of the tangent is

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

To compute the derivative, the following rules will need to be applied:

 AND

Product Rule:

Where u & v are differentiable functions.

Applying these rules:

We can factor out the common 4x and result in the following:

This is one of the answer choices.

To compute the derivative, understand that the derivative of the natural logarithm is found by:

Applying this results in:

After some rearranging:

Notice that the cosine & sine function all the same elements within the parenthesis. We can use the following identity to rewrite this function:

The function can now be rewritten as:

This is one of the answer choices.

In order to determine if this function is differential, we must see if the functions have the same values at  and if they have the same derivative at . In other words we have to determine if the function is continuous and differentiable. 

To determine if the functions are continuous, we must see if both piecewise functions are equal at . 

If we say  and , 

 ,

Therefore, the piecewise function is continuous at . 

To determine if it is differentiable, let's solve for the derivatives of both pieces at . 

 via the power rule.  via trigonometric identities. 

Solving the derivatives at x=1, 

, and . 

Since the functions do not have the same derivatives at , they are not differentiable. 

Therefore, the piecewise function is not a differential function because it isn't differentiable at 

We are given:

This particular function will use the chain rule and trigonometric rule for cosine and sine.

The problem requires us to take the derivative twice, and so we will proceed like so:

Taking the derivative of this new function will require the use of the product rule:

This is one of the answer choices.

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative: 

Quotient rule: 

Note that u and v may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

 at 

x:

y:

The slope is