Calculus I — Derivatives - Math

To find the concavity, we need to look at the first and second derivatives at the given point.

To take the first derivative of this equation, use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent:

Simplify:

Remember that anything to the zero power is equal to one.

The first derivative tells us if the function is increasing or decreasing. Plug in the given point, , to see if the result is positive (i.e. increasing) or negative (i.e. decreasing).

Therefore the function is increasing.

To find out if the function is convex, we need to look at the second derivative evaluated at the same point, , and check if it is positive or negative.

We're going to treat as since anything to the zero power is equal to one.

Notice that since anything times zero is zero.

Plug in our given value:

Since the second derivative is positive, the function is convex.

Therefore, we are looking at a graph that is both increasing and convex at our given point.

To find out if the function is increasing or decreasing, we need to look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

Anything to the zero power is one.

Now we plug in our given value and find out if the result is positive or negative. If it is positive, the function is increasing. If it is negative, the function is decreasing.

Therefore, the function is decreasing.

To find out if it is concave or convex, look at the second derivative. If the result is positive, it is convex. If it is negative, then it is concave.

To find the second derivative, we repeat the process using as our expression.

We're going to treat as .

Notice that since anything times zero is zero.

As stated before, anything to the zero power is one.

Since we get a positive constant, it doesn't matter where we look on the graph, as our second derivative will always be positive. That means that this graph is going to be convex at our given point.

Therefore, the function is decreasing and convex at our given point.

First, let's find out if the graph is increasing or decreasing. For that, we need the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as since anything to the zero power is one.

Notice that since anything times zero is zero.

Plug in our given point for . If the result is positive, the function is increasing. If the result is negative, the function is decreasing.

Our result is negative, therefore the function is decreasing.

To find the concavity, look at the second derivative. If the function is positive at our given point, it is concave. If the function is negative, it is convex.

To find the second derivative we repeat the process, but using as our expression.

As you can see, our second derivative is a constant. It doesn't matter what point we plug in for ; our output will always be negative. Therefore our graph will always be convex.

Combine our two pieces of information to see that at the given point, the graph is decreasing and convex.

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.

This question asks us to examine the concavity of the function . We will need to find the second derivative in order to determine where the function is concave upward and downward. Whenever its second derivative is positive, a function is concave upward.

Let us begin by finding the first derivative of f(x). We will need to use the Product Rule. According to the Product Rule, if , then . In this particular problem, let and . Applying the Product rule, we get

In order to evaluate the derivative of , we will need to invoke the Chain Rule. According to the Chain Rule, the derivative of a function in the form is given by . In finding the derivative of , we will let and .

We can now finish finding the derivative of the original function.

To summarize, the first derivative of the funciton is .

We need the second derivative in order to examine the concavity of f(x), so we will differentiate one more time. Once again, we will have to use the Product Rule in conjunction with the Chain Rule.

In order to find where f(x) is concave upward, we must find where f''(x) > 0.

In order to solve this inequality, we can divide both sides by . Notice that is always positive (because e raised to any power will be positive); this means that when we divide both sides of the inequality by , we won't have to flip the sign. (If we divide an inequality by a negative quantity, the sign flips.)

Dividing both sides of the inequality by gives us

When solving inequalities with polynomials, we often need to factor.

Notice now that the expression will always be positive, because the smallest value it can take on is 3, when x is equal to zero. Thus, we can safely divide both sides of the inequality by without having to change the direction of the sign. This leaves us with the inequality

, which clearly only holds when .

Thus, the second derivative of f''(x) will be positive (and f(x) will be concave up) only when . To represent this using interval notation (as the answer choices specify) we would write this as .

The answer is .

Since this function is a polynomial, we take the derivative of each term separately.

From the power rule, the derivative of

is simply

We can rewrite as

and using the power rule again, we get a derivative of

or

So the answer is

The question is just asking for the Quotient Rule formula.

Recall the Quotient Rule is the bottom function times the derivative of the top minus the top function times the derivative of the bottom all divided by the bottom function squared.

Given,

the bottom function is and the top function is . This makes the bottom derivative and the top derivative .

Substituting these into the Quotient Rule formula resulting in the following.

The chain rule is "first times the derivative of the second plus second times derivative of the first".

In this case, that means .

Since we're adding terms, we take the derivative of each part separately. For , we can use the power rule, which states that we multiply the variable by the current exponent and then lower the exponent by one. For sine, we use our trigonometric derivative rules.

Remember, .

To find the second derivative, we need to start by finding the first one.

Since we're adding terms, we take the derivative of each part separately. For , we can use the power rule, which states that we multiply the variable by the current exponent and then lower the exponent by one. For sine, we use our trigonometric derivative rules.

Remember, .

Now we repeat the process, but using as our equation.

It would be difficult to differentiate this equation by isolating . Luckily, we don't have to. Use to represent the derivative of with respect to and follow the chain rule.

(Remember, is the derivative of with respect to , although it usually doesn't get written out because it is equal to 1. We'll write it out this time so you can see how implicit differentiation works.)

Now we need to isolate by first putting all of these terms on the same side:

This is the equation for the derivative at any point on the curve. By substituting in (3, 2) from the original question, we can find the slope at that particular point:

For this problem, we can use the power rule. The power rule states that we multiply each variable by its current exponent and then lower that exponent by one.

Simplify.

Anything to the zero power is one, so .

Therefore, .

We use the power rule on each term of the function.

The first term

becomes

.

The second term

becomes

.

The final term, 7, is a constant, so its derivative is simply zero.

The average rate of change of on interval is

Substitute:

To get , we can use the power rule.

Since the exponent of the is , as , we lower the exponent by one and then multiply the coefficient by that original exponent:

Anything to the power is .

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

Remember that anything to the zero power is one.

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat as , as anything to the zero power is one.

That means this problem will look like this:

Notice that , as anything times zero is zero.

Remember, anything to the zero power is one.

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat as , as anything to the zero power is one.

Notice that , as anything times zero is zero.

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

Notice that since anything times zero is zero.

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

Notice that since anything times zero is zero.

That leaves us with .

Simplify.

As stated earlier, anything to the zero power is one, leaving us with: