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):

Trigonometric derivative: 

Product rule:  

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

Taking the derivative of the function  at the point 

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: 

Product rule: 

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

Taking the derivative of the function

The slope of the tangent is  at the point 

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):

Trigonometric derivative: 

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

Taking the derivative of the function  at the point 

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: 

Trigonometric derivative: 

Product rule: 

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

Taking the derivative of the function function  at the point 

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):

Trigonometric derivative: 

Product rule: 

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

Taking the derivative of the function  at the point 

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):

Quotient rule: 

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

Taking the derivative of the function  at the point 

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):

Trigonometric derivative: 

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

Taking the derivative of the function  at the point 

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):

Trigonometric derivative: 

Product rule: 

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

Taking the derivative of the function  at the point 

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):

Trigonometric derivative: 

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

Taking the derivative of the function  at the point 

The slope of the tangent is

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

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 are two solutions, though only one of them needs to fall within the interval  to satisfy the mean value theorem.

 does.