Applications of Derivatives - AP Calculus AB

Use the power rule on each term of the polynomial to get the derivative,

Now we plug in

We need to find the first derivative of f(x). This will require us to apply both the Product and Chain Rules. When we apply the Product Rule, we obtain:

In order to find the derivative of , we will need to employ the Chain Rule.

We can factor out a 2x to make this a little nicer to look at.

Now we must evaluate the derivative when x = .

The answer is .

To find the derivative of y with respect to x, we must take the differentiate both sides of the equation with respect to x:

The following rules were used:

, , , ,

Note that the chain rule was used everywhere we took the derivative of a function containing y, as well as in the exponential function. The product rule was used because both x and y are both functions of x.

Using algebra to solve, we get

To find the derivative with respect to y, we use implicit differentiation, which is an application of the chain rule

To answer this question, one first needs to find and then find the critical points of the function (i.e. where . Finally, one would need to determine the sign of for the intervals between the critical points.

For the given function:

.

Therefore, when and . So, the intervals to consider are:

To determine the sign of , pick any number for the given interval and evaluate at that number.

Therefore, is increasing on the intervals and since is greater than zero on these intervals.

To find the total distance traveled, the velocity function has to be integrated from to hours:

Finally, the question is asking how far the car will be from home. It was 3 miles from home when , so at , it will be:

miles from home.

To find if the equation is increasing or decreasing, we need to look at the first derivative. If our result is positive at , then the function is increasing. If it is negative, then the function is decreasing.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

Remember that anything to the zero power is one.

Plug in our given value.

Is it positive? Yes. Then it is increasing.

To find the concavity, we need to look at the second derivative. If it is positive, then the function is concave up. If it is negative, then the function is concave down.

Repeat the process we used for the first derivative, but use as our expression.

For this problem, we're going to say that since, as stated before, anything to the zero power is one.

Notice that as anything times zero is zero.

As you can see, there is no place for a variable here. It doesn't matter what point we look at, the answer will always be positive. Therefore this graph is always concave up.

This means that at our given point, the graph is increasing and concave up.

To find the intervals where the function is increasing, we need to find the points at which its slope changes from positive to negative and vice versa. The first derivative, which is the slope at any point, will help us.

First, we find the derivative of , using the power rule for each term. Recall that the power rule says

Also, the constant multiple rule will apply to the coefficients of each term. The constant multiple rule simply says that any constant factor of a term will "carry" to the derivative of that term. For example:

Lastly, the derivative of a constant is zero. This will result in the last term, , dropping off as we take the derivative.

Applying these rules, we find the derivative

Where the derivative is positive (blue line in graph), the tangent line to the original function is angled up. Where the derivative is negative (red line in graph), the slope of the tangent line is angled down.

The points where the tangent line's slope transitions from negative to positive or vice versa, are called the critical points. At these points, the tangent line becomes a horizontal line with a slope of zero (green line in graph). In other words, the "critical points" occur when the derivative is zero. These points will be the endpoints of our intervals of increasing and decreasing. To find the critical points, we will set the derivative equal to zero and solve for x. In this problem factoring is the best method:

Now that we have found the critical points, we need to know whether the original function is increasing or decreasing in the intervals between them. We will do so by testing a point in each interval and determining whether the derivative is positive or negative at that point. This is called the first derivative line test.

For the interval , we will test .

(Note: I will use the factored form of the derivative, but we could also use the polynomial version. Both will give the same result)

Since the derivative is negative at this point, we can conclude that the derivative is negative for the whole interval. Thus, the original function is decreasing on .

For the interval , we will test .

Since the derivative is positive at this point, we can again conclude that the derivative is positive for the whole interval. Thus, the original function is increasing on .

For the interval , we will test .

Since the derivative is negative at this point, we can again conclude that the derivative is negative for the whole interval. Thus, the original function is decreasing on .

Lastly, for the interval , we will test .

Since the derivative is positive at this point, we know that the derivative is positive for the whole interval. Thus the original function is increasing on .

From these 4 results, we now know the answer. The function is increasing on the intervals .

To find the coordinates of the local extrema of a function, we need to find the critical points of its first derivative.

Since is a polynomial, we can find its derivative term by term. The first 3 terms can be differentiated using the power rule, , and the constant multiple rule, .

The last term is a constant, and its derivative is zero.

Applying these rules, we find the first derivative:

Now we need to find the critical points. To do this, we set the first derivative equal to zero and solve. Factoring is the best method in this problem.

Now that we have the critical points, we need to determine for each one, whether it is a maximum, minimum, or neither. We use the first derivative line test to determine this.

For , we will test the interval before it, , and the interval after it, , and find whether they are increasing or decreasing.

For the interval , we will test and find whether is positive or negative.

Since is positive, the original function is increasing before the critical point, .

Now we will test the interval after .

For the interval, , we will test and find whether is positive or negative.

Since is negative, the original function is decreasing in the interval following .

Since the function is increasing before , and decreasing afterward, we can conclude that a maximum occurs at .

Now we find the value of this maximum, .

Thus is a local maximum.

Now we will determine whether a maximum or minimum occurs at .

We know that is decreasing before , but we still need to determine what happens afterward.

For the interval , we will test , and find whether is positive or negative.

Since is positive, the original function is increasing following .

Since is decreasing before, and increasing after , we can conclude that a minimum occurs at .

Now we need to find the value of this minimum, .

Thus is a local minimum.

So our answer is:

is a local maximum.

is a local minimum.

Concavity refers to the "curving" of the function. While the first derivative describes when the function is increasing or decreasing (instantaneous rate of change of ), the second derivative describes concavity, the instantaneous rate of change of .

The first derivative is like velocity, (moving forward or backward), while the second derivative is like acceleration (speeding up or slowing down).

Since we need to find the intervals of concavity, we will find the second derivative and work with it.

First we must find the first derivative using the power rule and constant multiple rule for each term of .

This gives:

Now to find the second derivative, we take the derivative of .

The same derivative rules apply:

Now that we have the second derivative, , we must find its critical points. We do this by setting , and solving.

The best method for this case is factoring.

The greatest common factor is ,

Inside the parentheses is a quadratic expression that can be factored like so:

Setting each factor equal to zero we find the following:

Now we know the critical points for the second derivative. To find the intervals of concavity, we test a point in each interval around these critical points and find whether the second derivative is positive or negative in that interval.

For the interval , we can test .

Using the factored form of the second derivative is easier than using the polynomial form, since the arithmetic involves fewer large numbers.

Since is negative, the original function is Concave Down in the interval .

Now for the interval , we can test .

We will combine the fractions inside the parentheses by getting the common denominator.

Multiplying the fractions gives:

Since is positive, the original function is Concave Up on the interval .

For the interval , we can test .

Since is negative, the original function is Concave Down on the interval .

Finally, for the interval , we can test .

Since is positive, the original function is Concave Up on the interval .

Summarizing the results, the intervals of concavity are:

Concave Down:

Concave Up:

On the domain , we know that the derivative begins positive, and because the concavity is positive, we know the derivative is increasing. Thus, the derivative stays positive for this entire interval, and the function increases from 2 to 4. Thus, must be greater than .

In the case of the interval , we know that the derivative is increasing, but it starts out negative. Thus, perhaps the derivative only increased from -1 to -0.5 in this interval, and the function would have decreased the entire time. In this case, would be less than , so we can't really say anything about these values.

For the remaining two, there's not any clear way to relate the functions at and . While we know needs to be bigger than 1, we don't know by how much. Similarly, while we know needs to be bigger than -1, we don't know by how much. Thus, it's completely possible that and .

As for and , we know even less. In between the two intervals, our function could have shot up a million, or shot down by the same amount. Thus, there's no safe comparison we can make between these two values.

We are asked to find the normal line. This means we need to find the line that is perpendicular to the tangent line at . In order to find the tangent line, we will need to evaluate the derivative of at .

The slope of the tangent line at is . Because the tangent line and the normal line are perpendicular, the product of their slopes must equal .

(slope of tangent)(slope of normal) =

We now have the slope of the normal line. Once we find a point through which it passes, we will have enough information to derive its equation.

Since the normal line passes through the function at , it will pass through the point . Be careful to use the original equation for , not its derivative.

The normal line has a slope of and passes through the piont . We can now use point-slope form to find the equation of the normal line.

Multiply both sides by .

The answer is .

We simply differentiate both sides of the equation

(Don't forget the chain rule)

Now we solve for

Implicit differentiation is very similar to normal differentiation, but every time we take a derivative with respect t , we need to multiply the result by We also differentiate the entire equation from left to right, including any numbers. Then, we solve for that for our final answer.

Implicit differentiation requires taking the derivative of everything in our equation, including all variables and numbers. Any time we take a derivative of a function with respect to , we need to implicitly write after it. Hence, the name of this method. Then, we solve for

To determine the rate of the change of the angle opposite to the base of the given right triangle, we must relate it to the rate of change of the base of the triangle when the triangle is a certain area.

First we must determine the length of the base of the right triangle at the given area:

Now, we must find something that relates the angle opposite of the base to the length of the base and height - the tangent of the angle:

To find the rate of change of the angle, we take the derivative of both sides with respect to time, keeping in mind that the base of the triangle is dependent on time, while the height is constant:

We know the rate of change of the base, and we can find the angle from the sides of the triangle:

Plugging this and the other known information in and solving for the rate of change of the angle adjacent to the base, we get

radians per minute

To find , we must take the derivative of both sides of the equation with respect to x. When we do this, we get

The derivatives were found using the following rules:

, , ,

Note that for every derivative of a function of y with respect to x, the chain rule was used, which accounts for appearing.

Algebraic simplification gets us our final answer,

To find the derivative of the function, we must use implicit differentiation, which is an application of the chain rule. We start by taking the derivative of the function with respect to x, noting that whenever we take a derivative of y, it is with respect to x, so we denote it as .

Bringing the terms with to one side and factoring it out, we get

To find the derivative of the function, we must use implicit differentiation, which is an application of the chain rule. We start by taking the derivative of the function with respect to x, noting that whenever we take a derivative of y, it is with respect to x, so we denote it as .

Bringing the terms with to one side and factoring it out, we get