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.

 which satisfies the MVT by falling within 

It may seem daunting, but just treat  as another constant.

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: 

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

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.

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

Derivative of an exponential: 

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

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.

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

Derivative of an exponential: 

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

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 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:

Derivative of an exponential: 

Note that u 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.

Taking the partial derivatives of  at 

:

:

The slope is 

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):

Note that  may represent complext functions.

Taking the derivative of the function   at 

The slope of the tangent is

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: 

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

Taking the derivative of the function  at 

The slope of the tangent is

First, find the derivative using the power rule.

Remember the power rule:

We can now apply this to our situation.

The derivative is 

Now, substitute  for .

First, find the derivative.

Remember the power rule:

We can now apply this to our situation.

The derivative is .

Now, substitute  for . 

First, find the derivative using the power rule.

Remember the power rule:

We can now apply this to our situation.

Recall that the derivative of a constant is zero.

Thus, the derivative is 

Now, substitute  for .