Derivative Rules for Sums, Products, and Quotients - AP Calculus BC
Card 0 of 200
Find the second derivative of g(x)

Find the second derivative of g(x)
Tap to see back →
Find the second derivative of g(x)

To find this derivative, we need to use the product rule:

So, let's begin:


So, we are closer, but we need to derive again to get the 2nd derivative


So, our answer is:

Find the second derivative of g(x)
To find this derivative, we need to use the product rule:
So, let's begin:
So, we are closer, but we need to derive again to get the 2nd derivative
So, our answer is:
Suppose
and
are differentiable functions, and
. Calculate the derivative of
, at 
Suppose and
are differentiable functions, and
. Calculate the derivative of
, at
Tap to see back →
The correct answer is 11.
Taking the derivative of
involves the product rule, and the chain rule.
![h'(x)=[f(x+2)][-g'(-x)]+[g(-x)][f'(x+2)]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/250996/gif.latex)

Substituting
into both sides of the derivative we get
.
The correct answer is 11.
Taking the derivative of involves the product rule, and the chain rule.
Substituting into both sides of the derivative we get
.

What is the slope of
at
?
What is the slope of at
?
Tap to see back →
In order to find the slope of a function at a certain point, plug in that point into the first derivative of the function. Our first step here is to take the first derivative.
Since we see that f(x) is composed of two different functions, we must use the product rule. Remember that the product rule goes as follows:


Following that procedure, we set
equal to
and
equal to
.
,
which can be simplified to
.
Now plug in 1 to find the slope at x=1.

Remember that
.
In order to find the slope of a function at a certain point, plug in that point into the first derivative of the function. Our first step here is to take the first derivative.
Since we see that f(x) is composed of two different functions, we must use the product rule. Remember that the product rule goes as follows:
Following that procedure, we set equal to
and
equal to
.
,
which can be simplified to
.
Now plug in 1 to find the slope at x=1.
Remember that .
Find the value of the derivate of the given function at the point
:

Find the value of the derivate of the given function at the point :
Tap to see back →
To solve this problem, first we need to take the derivative of the function. To do this we need to use the quotient rule and simplified as follows:

From here we need to evaluate at the given point
. In this case, only the x value is important, so we evaluate our derivative at x=2 to get
.
To solve this problem, first we need to take the derivative of the function. To do this we need to use the quotient rule and simplified as follows:
From here we need to evaluate at the given point . In this case, only the x value is important, so we evaluate our derivative at x=2 to get
.
Find the second derivative of the given function:

Find the second derivative of the given function:
Tap to see back →
To find the second derivative, first we need to find the first derivative. To find the first derivative we need to use the quotient rule as follows. So for the given function, we get the first derivative to be
.
Now we have to take the derivative of the derivative. To do this we need to use the quotient rule as shown below.
Thus, we get

To find the second derivative, first we need to find the first derivative. To find the first derivative we need to use the quotient rule as follows. So for the given function, we get the first derivative to be
.
Now we have to take the derivative of the derivative. To do this we need to use the quotient rule as shown below.
Thus, we get
Given:

Find f'(x):
Given:
Find f'(x):
Tap to see back →
Computation of the derivative requires the use of the Product Rule and Chain Rule.
The Product Rule is used in a scenario when one has two differentiable functions multiplied by each other:

![$\frac{d}{dx}$[f(x)*g(x)]=f(x)*g'(x)+g(x)*f(x)](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/400721/gif.latex)
This can be easily stated in words as: "First times the derivative of the second, plus the second times the derivative of the first."
In the problem statement, we are given:

is the "First" function, and
is the "Second" function.
The "Second" function requires use of the Chain Rule.
When:

Applying these formulas results in:

Simplifying the terms inside the brackets results in:

We notice that there is a common term that can be factored out in the sets of equations on either side of the "+" sign. Let's factor these out, and make the equation look "cleaner".

Inside the brackets, it is possible to clean up the terms into one expanded function. Let us do this:

Simplifying this results in one of the answer choices:

Computation of the derivative requires the use of the Product Rule and Chain Rule.
The Product Rule is used in a scenario when one has two differentiable functions multiplied by each other:
This can be easily stated in words as: "First times the derivative of the second, plus the second times the derivative of the first."
In the problem statement, we are given:
is the "First" function, and
is the "Second" function.
The "Second" function requires use of the Chain Rule.
When:
Applying these formulas results in:
Simplifying the terms inside the brackets results in:
We notice that there is a common term that can be factored out in the sets of equations on either side of the "+" sign. Let's factor these out, and make the equation look "cleaner".
Inside the brackets, it is possible to clean up the terms into one expanded function. Let us do this:
Simplifying this results in one of the answer choices:
Find the slope of the tangent line to the function
at
.
Find the slope of the tangent line to the function at
.
Tap to see back →
The slope of the tangent line to a function at a point is the value of the derivative of the function at that point. In this problem,
is a quotient of two functions,
, so the quotient rule is needed.
In general, the quotient rule is
.
To apply the quotient rule in this example, you must also know that
and that
.
Therefore, the derivative is

The last step is to substitute
for
in the derivative, which will tell us the slope of the tangent line to
at
.

The slope of the tangent line to a function at a point is the value of the derivative of the function at that point. In this problem, is a quotient of two functions,
, so the quotient rule is needed.
In general, the quotient rule is
.
To apply the quotient rule in this example, you must also know that and that
.
Therefore, the derivative is
The last step is to substitute for
in the derivative, which will tell us the slope of the tangent line to
at
.
Tap to see back →
Evaluation of this integral requires use of the Product Rule. One must also need to recall the form of the derivative of
.
Product Rule:
![$\frac{d}{dx}$[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formulaimage/image/430297/gif.latex)

Applying these two rules results in:



This matches one of the answer choices.
Evaluation of this integral requires use of the Product Rule. One must also need to recall the form of the derivative of .
Product Rule:
Applying these two rules results in:
This matches one of the answer choices.
Compute the derivative:

Compute the derivative:
Tap to see back →
Computation of this derivative will require the use of the Product Rule, and knowledge of the derivative of the inverse tangent function, and natural logarithmic function:

![$\frac{d}{dx}$[\ln(u)]=\frac{1}{u}$$\frac{du}{dx}$](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/470324/gif.latex)
We can now easily compute the derivative.

This simplifies to:

This is one of the answer choices.
Computation of this derivative will require the use of the Product Rule, and knowledge of the derivative of the inverse tangent function, and natural logarithmic function:
We can now easily compute the derivative.
This simplifies to:
This is one of the answer choices.
Find dy/dx:

Find dy/dx:
Tap to see back →
Solving for the derivative requires knowledge of the rule for the inverse tangent function:

In our case:

We can take the derivative of this using the product rule:


Now we can simply plug all of this into the above formula and we arrive at:

Simplifying this further gives:

Solving for the derivative requires knowledge of the rule for the inverse tangent function:
In our case:
We can take the derivative of this using the product rule:
Now we can simply plug all of this into the above formula and we arrive at:
Simplifying this further gives:
Find the derivative of the function

Find the derivative of the function
Tap to see back →
We find the answer using the quotient rule
![($\frac{f(x)}{g(x)}$)' = $$\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2$}$](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/592262/gif.latex)
and the product rule

and then simplifying.



or
. The extra brackets in the denominator are optional.
We find the answer using the quotient rule
and the product rule
and then simplifying.
or
. The extra brackets in the denominator are optional.
If
, find
in terms of
and
.
If , find
in terms of
and
.
Tap to see back →
Using a combination of logarithms, implicit differentiation, and a bit of algebra, we have




. Quotient Rule + implicit differentiation.






Using a combination of logarithms, implicit differentiation, and a bit of algebra, we have
. Quotient Rule + implicit differentiation.
Find the derivative of the function 
Find the derivative of the function
Tap to see back →
The correct answer is
.
Using the Quotient Rule and the fact
, we have





.
The correct answer is .
Using the Quotient Rule and the fact , we have
.
Tap to see back →
First, factor out
:
. Now we can differentiate using the product rule,
.
Here,
so
.
so
.
The answer is 
First, factor out :
. Now we can differentiate using the product rule,
.
Here, so
.
so
.
The answer is
Tap to see back →
According to the product rule,
. Here
so
.
so
.
The derivative is 
Factoring out the 2 gives
. Remembering the double angle trigonometric identity finally gives 
According to the product rule, . Here
so
.
so
.
The derivative is
Factoring out the 2 gives . Remembering the double angle trigonometric identity finally gives
If
, find 
If , find
Tap to see back →
First, we need to find
. We can do that by using the quotient rule.
.
Plugging
in for
and simplifying, we get
.
First, we need to find . We can do that by using the quotient rule.
.
Plugging in for
and simplifying, we get
.
Find the derivative of f:

Find the derivative of f:
Tap to see back →
The derivative of the function is equal to

and was found using the following rules:
,
,
, 
The derivative of the function is equal to
and was found using the following rules:
,
,
,
Find the derivative of the function:

where
is a constant
Find the derivative of the function:
where is a constant
Tap to see back →
When taking the derivative of the sum, we simply take the derivative of each component.
The derivative of the function is

and was found using the following rules:
, 
When taking the derivative of the sum, we simply take the derivative of each component.
The derivative of the function is
and was found using the following rules:
,
Compute the first derivative of the following function.

Compute the first derivative of the following function.
Tap to see back →
Compute the first derivative of the following function.

To solve this problem, we need to apply the product rule:

So, we need to apply this rule to each of the terms in our function. Let's start with the first term

Next, let's tackle the second part

Now, combine the two to get:

Compute the first derivative of the following function.
To solve this problem, we need to apply the product rule:
So, we need to apply this rule to each of the terms in our function. Let's start with the first term
Next, let's tackle the second part
Now, combine the two to get:
Evaluate the derivative of the function
.
Evaluate the derivative of the function .
Tap to see back →
Use the product rule: 
where
and
.
By the power rule,
.
By the chain rule,
.
Therefore, the derivative of the entire function is:

.
Use the product rule:
where and
.
By the power rule, .
By the chain rule, .
Therefore, the derivative of the entire function is:
.