AP Calculus AB › Differential Functions
Find the derivative:
If , then the derivative is
.
If , the the derivative is
.
If , then the derivative is
.
If , then the derivative is
.
If , then the derivative is
.
There are many other rules for the derivatives for trig functions.
If , then the derivative is
. This is known as the chain rule.
In this case, we must find the derivative of the following:
That is done by doing the following:
Therefore, the answer is:
What is the slope of the function at the point
?
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 rule will be necessary:
Derivative of an exponential:
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.
Evaluating the derivatives of at the point
x:
y:
Thus the slope is
What is the slope of the line normal to the function at the point
?
The first step to finding the slope of the line normal to a point is to find the slope of the tangent at this point.
The slope of this tangent, in turn, is found by finding the value of the derivative of the function at this point.
Evaluating the function at the point
The slope of the tangent is
The slope of the normal line is the negative reciprocal of this value. Thus for this problem the normal is
What is the derivative of the function ?
Note that for this problem, we're not told to take the derivative with respect to any particular variable, so it would be prudent to take the derivative with respect to all three. Knowledge of the following derivative rules will be necessary:
Derivative of an exponential:
Trigonometric derivative:
Product rule:
Note that and
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.
Looking at our function, take the derivative with respect to each variable
The complete derivative is then the sum of these:
Find the derivative of the function.
To find the derivative, use the power rule which states, .
Applying the power rule to each term in the function we get,
.
Recall that the derivative of a constant is 0.
Thus, the derivative is:
Which of the following is an inflection point of ?
The points of inflection of a function occur where the second derivative of the funtion is equal to zero.
Find this second derivative by taking the derivative of the function twice:
Set the second derivative to zero and find the values that satisfy the equation:
Now, plug these values back in to the original function to find the values of the function that match to them:
The two points of inflection are
can be shown to be to be a point of inflection by observing the sign change at lower and higher values
What is the slope of the tangent line at ?
In order to find the slope of the tangent line at , find the derivative of the function.
Because the derivative is a constant, the slope of the tangent line is also constant: .
Approximate the integral of for
to
using midpoint Reimann sums and three midpoints.
The form of a Riemann sum follows:
The interval can be divided into three intervals of length
, with midpoints of
.
Therefore, the approximation of the integral is:
Using the method of midpoint Reimann sums, approximate the integral of the function over the interval
using four midpoints.
The Reimann sum approximation of an integral of a function with subintervals over an interval
takes the form:
Where is the length of the subintervals.
For this problem, since there are four midpoints, the subintervals have length , and the midpoints are
.
The integral is thus:
Use Riemann midpoint sums to approximate the area between the curve and the
-axis between
and
. Use five intervals
.
We are looking to approximate the area between 0 and 3, and we are dividing it into 5 intervals, so each interval will have a length of .
The midpoints of the 5 intervals are at , so we'll be evaluating the function for these values of x.
The formula for the Riemann midpoint sum is .
Note that you get the same answer regardless if you multiply each individual times
or calculate the sum first then multiply by 0.6.
In this case, adding together the heights at the midpoints and multiplying by 0.6 yields .