Derivative rules for sums, products, and quotients of functions - AP Calculus AB

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.

Differentiate both sides and proceed with the product rule:

(1)

Evaluate the derivatives in each term. For the first term,

(2)

apply the chain rule,

So now the first term in equation (2) can be written,

(3)

The second term in equation (2) is easy, this is just the product of multiplied by the derivative of ,

(4)

Combine equations (3) and (4) to write the derivative,

Use the product rule to find the derivative.

Use the power rule to find the derivative.

Thus, the derivative is

Here we use the product rule:

Let and

Then (using the chain rule)

and (using the chain rule)

Subbing these values back into our equation gives us

Simplify by combining like-terms

and pulling out a from each term gives our final answer

To find the derivative of the function, we use the quotient rule, which is

, where and are any expression

Using the function from the problem statement, we get

Taking the derivatives, we get

When evaluating the derivative, pay attention to the fact that are constants, (not variables) and are treated as such.

.

and hence

.

To obtain an expression for , we can take the derivative of using the sum rule.

.

Substituting into this equation gives us

.

To find , we will need to use the quotient rule; .

. Start

. Use the quotient rule.

. Take the derivatives inside of the quotient rule. The derivative of uses the product rule.

. Simplify to match the correct answer.

Given that there are 2 terms in the numerator and only one in the denominator, one can split up the equation into 2 separate derivatives:

.

Now we simplify these, and proceed to solve:

Because this problem contains two functions multiplied together that can't be simplified any further, it calls for the product rule, which states that .

By using this rule, we get the answer:

By simplifying, we conclude that the derivative is equal to

.

Because this problem contains two functions multiplied together that can't be simplified any further, it calls for the product rule, which states that

By applying this rule to the equation

we get

Because we are dealing with a quotient that cannot be simplified, we use the quotient rule, which states that if

,

.

By observing the given equation

,

we can see that

and

.

Therefore, the derivative is

.

Because we are differentiating a quotient that cannot be simplified, we must use the quotient rule, which states that if

,

then

.

By observing the given equation,

,

we see that in this case,

and

.

Given this information, the quotient rule tells us that

.

This problem is a quotient rule inside of a chain rule. First, let's look at the chain rule:

.

Given this, we can deduce that since

,

and

.

By plugging these into the chain rule formula, we get

To find the derivative of the

second term, we must use the quotient rule, which states that the the derivative

of a quotient is ((denominator)(derivative of numerator)-(numerator)(derivative

of denominator))/(denominator squared). Using this rules we find that

.

By plugging this back in, we find the final derivative to be

First notice that the function is a product of two functions and . Apply the product rule:

                                                                                                                                 The second term will require the chain rule. Recall that the derivative for a radical function-of-a-function is given by: